# 6. If you 5.5 mol of Nitrogen in: N2 + 3H2 → 2NH3a. How many moles of Hydrogen are needed to completely react with the nitrogen?b. How many moles

6. If you 5.5 mol of Nitrogen in: N2 + 3H2 → 2NH3 a. How many moles of Hydrogen are needed to completely react with the nitrogen?

b. How many moles of Ammonia will be produced?

## This Post Has 9 Comments

1. Pordie says:

Hope this helps you.
$N2 + 3h2 > 2nh3 we want to produce 2.75 mol of nh3. how many moles of nitrogen would be required$
$N2 + 3h2 > 2nh3 we want to produce 2.75 mol of nh3. how many moles of nitrogen would be required$

1)    2 HgO    →      2 Hg + O₂

(2 x 216.59 g) → gives → (2 x 200.59 g)

433.18 g   → gives → 401.18 g

8.74 g     → gives → ? g

by cross multiplication:

Theoretical yield = (401.18 x 8.74) / 433.18 = 8.094 g

% yield = (actual yield / theoretical yield) * 100

= (6.42 / 8.09) * 100 = 79.3 %

2) F₂ + 2 NaBr → 2 NaF + Br₂

1 mole F₂ → gives → 2 moles NaF

0.24 mol F₂ will give ?? mol NaF

= (0.24 x 2) / 1 = 0.48 mol NaF

3) 4 Fe(s) + 3 O₂(g) → 2 Fe₂O₃(s)

number of moles of Fe = (54.6 / 55.845) = 0.97 mol

number of moles of O₂ = (91.6 g / 32) = 2.8625 mol

4 moles Fe react with 3 moles O₂

0.97 mol Fe will react completely with 0.7275 mol O₂

so 0.97 mol Fe is the limiting reactant and O₂ present in excess

4 moles Fe(s)    → gives → 2 moles Fe₂O₃

0.97 mol Fe(s)  → gives → ? moles Fe₂O₃

= (0.97 x 2) / 4 = 0.485 mol Fe₂O₃

mass of Fe₂O₃ = 0.485 mol * 159.69 g/mol = 77.45 g

4) Ga₂O₃(s) + 3 SOCl₂(l) → 2 GaCl₃(s) + 3 SO₂

Molar mass of Ga₂O₃ = 187.44 g/mol

Molar mass of SOCl₂ = 118.97 g/mol

Mass of Ga₂O₃ = 71.8 g

Mass of SOCl₂ = 110.8 g

number of moles = mass (g) / molar mass

number of moles of Ga₂O₃ = 71.8 / 187.44 = 0.38 mol

number of moles of SOCl₂ = 110.8 / 118.97 = 0.93 mol

1 mole Ga₂O₃(s) react with 3 moles SOCl₂

0.38 mol Ga₂O₃ react with 1.14 mole SOCl₂

But only 0.93 mol SOCl₂ present so it is the limiting reactant

3 moles of SOCl₂ → gives → 2 moles GaCl₃

0.93 mol SOCl₂  → gives → ? mole GaCl₃

= (2 x 0.93) / 3 = 0.62 mole

Mass of GaCl₃ = 0.62 * 176.073 = 109.16 g

5) N₂(g) + 3 H₂(g) → 2 NH₃(g)

1 mol N₂ reacts with 3 moles H₂

so 6 moles N₂ will react completely with 18 moles H₂

But moles of H₂ present is only 10 moles, so Nitrogen is considered as the excess reactant.

6) B₂O₃(s) + 3 C(s) + 3 Cl₂(g)   → 2 BCl₃(g) + 3 CO(g)

3 moles C → gives → 2 moles BCl₃

? moles C → gives → 3 moles BCl₃

moles of C = (3*3) / 2 = 4.5 moles C

Mass of carbon required = 4.5 moles * 12 g/mol = 54 g carbon

7) B₂O₃(s) + 3 C(s) + 3 Cl₂(g) → 2 BCl₃(g) + 3 CO(g)

number of moles of B₂O₃ =  mass / molar mass

= 122 g / 69.6182 = 1.75 moles

1 mole B₂O₃       → gives → 2 moles BCl₃

1.75 mole B₂O₃  → gives → ? moles BCl₃

= 2 * 1.75 = 3.5 moles BCl₃

8) 4 Fe(s) + 3 O₂ → 2 Fe₂O₃

moles of Fe = (54.6 g) / (55.845 g/mol) = 0.98 mol

moles of O₂ = (91.6 g) / (32 g/mol) = 2.8625 mol

4 moles Fe(s) react with 3 moles O₂(g)

0.98 mol Fe(s) react completely with ? moles O₂(g)

= 0.735 mol O₂

moles of excess oxygen = 2.8625 - 0.735 = 2.1275 mol

mass of excess oxygen = 2.1275 * 32 = 68.08 g O₂

9) 2 NO(g) + O₂(g) → 2 NO₂

2 moles NO react with 1 mol O₂

? moles NO will react with 2.5 mol of O₂

= (2 * 2.5) / 1 = 5 moles NO

Mass of NO = 5 * 30 = 150 g

10) N₂(g) + 3 H₂(g) → 2 NH₃(g)

1 mole N₂ reacts with 3 moles H₂

so 3 moles of N₂ will react completely with 9 moles H₂

but only 5 moles of Hydrogen present so it is the limiting reactant

11) 2 HgO(s)        →      2 Hg(l) + O₂(g)

(2 x 216.59 g) give    (2 x 200.59)

433.18 g will give    401.18 g

8.74 g HgO  will give ? g Hg

Theoretical yield of Hg = (8.74 x 401.18) / 433.18 = 8.094 g

12) B₂O₃(s) + 3 C(s) + 3 Cl₂(g) → 2BCl₃(g) + 3 CO(g)

number of moles of BCl₃ produced = 175 g / 117.17 g/mol = 1.5 moles

3 moles of Cl₂ produce 2 moles BCl₃

? moles of Cl₂ produces 1.5 mole BCl₃

= (1.5 x 3) / 2 = 2.25 moles Cl₂

13) 4 Fe(s) + 3 O₂ → 2 Fe₂O₃

moles of Fe = (54.6 g) / (55.845 g/mol) = 0.98 mol

moles of O₂ = (91.6 g) / (32 g/mol) = 2.8625 mol

4 moles Fe(s) react with 3 moles O₂(g)

0.98 mol Fe(s) react completely with ? moles O₂(g)

= 0.735 mol O₂

So the limiting reactant is Fe

14) N₂(g) + 3 H₂(g) → 2 NH₃(g)

1 mole N₂ reacts with 3 moles H₂

so 6 moles of N₂ will react completely with 18 moles H₂

but only 10 moles of Hydrogen present so it is the limiting reactant

3 moles of H₂ gives 2 moles NH₃

10 moles of H₂ will give ? moles NH₃

= (10 * 2) / 3 = 6.67 moles NH₃

15) B₂O₃(s) + 3 C(s) + 3 Cl₂(g) → 2BCl₃(g) + 3 CO(g)

number of moles of BCl₃ produced = 700 g / 117.17 g/mol = 5.97 moles

3 moles of C produce 2 moles BCl₃

? moles of C produces 5.97 mole BCl₃

= (5.97 x 3) / 2 = 8.96 moles C

16) Fe₂O₃(s) + 3 CO(g) → 2 Fe(s) + 3CO₂(g)

1 mole of Fe₂O₃ gives 3 moles CO₂

? mole of Fe₂O₃ gives 7.4 moles CO₂

= 2.47 moles Fe₂O₃

3 moles of CO gives 3 moles of CO₂

so 7.4 moles of CO will produce the required amount of CO₂ (7.4)

3. edith47 says:

Therefore 9 moles of nitrogen are required to produce 18 moles of ammonia.

4. mayaa2351 says:

Moles NH₃: 0.0593

0.104 moles of N₂ remain

Final pressure: 0.163atm

Explanation:

The reaction of nitrogen with hydrogen to produce ammonia is:

N₂ + 3 H₂ → 2 NH₃

Using PV = nRT, moles of N₂ and H₂ are:

N₂: 1atmₓ3.0L / 0.082atmL/molKₓ273K = 0.134 moles of N₂

H₂: 1atmₓ2.0L / 0.082atmL/molKₓ273K = 0.089 moles of H₂

The complete reaction of N₂ requires:

0.134 moles of N₂ × (3 moles H₂ / 1 mole N₂) = 0.402 moles H₂

That means limiting reactant is H₂. And moles of NH₃ produced are:

0.089 moles of H₂ × (2 moles NH₃ / 3 mole H₂) = 0.0593 moles NH₃

Moles of N₂ remain are:

0.134 moles of N₂ - (0.089 moles of H₂ × (1 moles N₂ / 3 mole H₂)) = 0.104 moles of N₂

And final pressure is:

P = nRT / V

P = (0.104mol + 0.0593mol)×0.082atmL/molK×273K / 5.0L

P = 0.163atm

5. tishfaco5000 says:

2.75 mol

Explanation:

Given data:

Mass of Nitrogen = 38.5 g

Moles of ammonia produced = ?

Solution:

Chemical  equation:

N₂ + 3H₂    →     2NH₃

Number of moles of nitrogen:

Number of moles = mass/ molar mass

Number of moles = 38.5 g/ 28 g/mol

Number of moles = 1.375 mol

Now we will compare the moles of ammonia and nitrogen from balance chemical equation.

N₂            :            NH₃

1              :             2

1.375       :           2×1.375 = 2.75 mol

Thus 2.75 moles of ammonia  are produced from 38.5 g of nitrogen.

c

Explanation:

7. pooch868 says:

1)    2 HgO    →      2 Hg + O₂
(2 x 216.59 g) → gives → (2 x 200.59 g)
433.18 g   → gives → 401.18 g
8.74 g     → gives → ? g
by cross multiplication:
Theoretical yield = (401.18 x 8.74) / 433.18 = 8.094 g
% yield = (actual yield / theoretical yield) * 100
= (6.42 / 8.09) * 100 = 79.3 %

2) F₂ + 2 NaBr → 2 NaF + Br₂
1 mole F₂ → gives → 2 moles NaF
0.24 mol F₂ will give ?? mol NaF
= (0.24 x 2) / 1 = 0.48 mol NaF

3) 4 Fe(s) + 3 O₂(g) → 2 Fe₂O₃(s)
number of moles of Fe = (54.6 / 55.845) = 0.97 mol
number of moles of O₂ = (91.6 g / 32) = 2.8625 mol
4 moles Fe react with 3 moles O₂
0.97 mol Fe will react completely with 0.7275 mol O₂
so 0.97 mol Fe is the limiting reactant and O₂ present in excess
4 moles Fe(s)    → gives → 2 moles Fe₂O₃
0.97 mol Fe(s)  → gives → ? moles Fe₂O₃
= (0.97 x 2) / 4 = 0.485 mol Fe₂O₃
mass of Fe₂O₃ = 0.485 mol * 159.69 g/mol = 77.45 g

4) Ga₂O₃(s) + 3 SOCl₂(l) → 2 GaCl₃(s) + 3 SO₂
Molar mass of Ga₂O₃ = 187.44 g/mol
Molar mass of SOCl₂ = 118.97 g/mol
Mass of Ga₂O₃ = 71.8 g
Mass of SOCl₂ = 110.8 g
number of moles = mass (g) / molar mass
number of moles of Ga₂O₃ = 71.8 / 187.44 = 0.38 mol
number of moles of SOCl₂ = 110.8 / 118.97 = 0.93 mol
1 mole Ga₂O₃(s) react with 3 moles SOCl₂
0.38 mol Ga₂O₃ react with 1.14 mole SOCl₂
But only 0.93 mol SOCl₂ present so it is the limiting reactant
3 moles of SOCl₂ → gives → 2 moles GaCl₃
0.93 mol SOCl₂  → gives → ? mole GaCl₃
= (2 x 0.93) / 3 = 0.62 mole
Mass of GaCl₃ = 0.62 * 176.073 = 109.16 g

5) N₂(g) + 3 H₂(g) → 2 NH₃(g)
1 mol N₂ reacts with 3 moles H₂
so 6 moles N₂ will react completely with 18 moles H₂
But moles of H₂ present is only 10 moles, so Nitrogen is considered as the excess reactant.

6) B₂O₃(s) + 3 C(s) + 3 Cl₂(g)   → 2 BCl₃(g) + 3 CO(g)
3 moles C → gives → 2 moles BCl₃
? moles C → gives → 3 moles BCl₃
moles of C = (3*3) / 2 = 4.5 moles C
Mass of carbon required = 4.5 moles * 12 g/mol = 54 g carbon

7) B₂O₃(s) + 3 C(s) + 3 Cl₂(g) → 2 BCl₃(g) + 3 CO(g)
number of moles of B₂O₃ =  mass / molar mass
= 122 g / 69.6182 = 1.75 moles
1 mole B₂O₃       → gives → 2 moles BCl₃
1.75 mole B₂O₃  → gives → ? moles BCl₃
= 2 * 1.75 = 3.5 moles BCl₃

8) 4 Fe(s) + 3 O₂ → 2 Fe₂O₃
moles of Fe = (54.6 g) / (55.845 g/mol) = 0.98 mol
moles of O₂ = (91.6 g) / (32 g/mol) = 2.8625 mol
4 moles Fe(s) react with 3 moles O₂(g)
0.98 mol Fe(s) react completely with ? moles O₂(g)
= 0.735 mol O₂
moles of excess oxygen = 2.8625 - 0.735 = 2.1275 mol
mass of excess oxygen = 2.1275 * 32 = 68.08 g O₂

9) 2 NO(g) + O₂(g) → 2 NO₂
2 moles NO react with 1 mol O₂
? moles NO will react with 2.5 mol of O₂
= (2 * 2.5) / 1 = 5 moles NO
Mass of NO = 5 * 30 = 150 g

10) N₂(g) + 3 H₂(g) → 2 NH₃(g)
1 mole N₂ reacts with 3 moles H₂
so 3 moles of N₂ will react completely with 9 moles H₂
but only 5 moles of Hydrogen present so it is the limiting reactant

11) 2 HgO(s)        →      2 Hg(l) + O₂(g)
(2 x 216.59 g) give    (2 x 200.59)
433.18 g will give    401.18 g
8.74 g HgO  will give ? g Hg
Theoretical yield of Hg = (8.74 x 401.18) / 433.18 = 8.094 g

12) B₂O₃(s) + 3 C(s) + 3 Cl₂(g) → 2BCl₃(g) + 3 CO(g)
number of moles of BCl₃ produced = 175 g / 117.17 g/mol = 1.5 moles
3 moles of Cl₂ produce 2 moles BCl₃
? moles of Cl₂ produces 1.5 mole BCl₃
= (1.5 x 3) / 2 = 2.25 moles Cl₂

13) 4 Fe(s) + 3 O₂ → 2 Fe₂O₃
moles of Fe = (54.6 g) / (55.845 g/mol) = 0.98 mol
moles of O₂ = (91.6 g) / (32 g/mol) = 2.8625 mol
4 moles Fe(s) react with 3 moles O₂(g)
0.98 mol Fe(s) react completely with ? moles O₂(g)
= 0.735 mol O₂
So the limiting reactant is Fe

14) N₂(g) + 3 H₂(g) → 2 NH₃(g)
1 mole N₂ reacts with 3 moles H₂
so 6 moles of N₂ will react completely with 18 moles H₂
but only 10 moles of Hydrogen present so it is the limiting reactant
3 moles of H₂ gives 2 moles NH₃
10 moles of H₂ will give ? moles NH₃
= (10 * 2) / 3 = 6.67 moles NH₃

15) B₂O₃(s) + 3 C(s) + 3 Cl₂(g) → 2BCl₃(g) + 3 CO(g)
number of moles of BCl₃ produced = 700 g / 117.17 g/mol = 5.97 moles
3 moles of C produce 2 moles BCl₃
? moles of C produces 5.97 mole BCl₃
= (5.97 x 3) / 2 = 8.96 moles C

16) Fe₂O₃(s) + 3 CO(g) → 2 Fe(s) + 3CO₂(g)
1 mole of Fe₂O₃ gives 3 moles CO₂
? mole of Fe₂O₃ gives 7.4 moles CO₂
= 2.47 moles Fe₂O₃
3 moles of CO gives 3 moles of CO₂
so 7.4 moles of CO will produce the required amount of CO₂ (7.4)

8. hardwick744 says:

c.

Explanation:

The difference between the two is the measure of pressure. Atm (atmospheres) or kPa (kilopascals) both are measures of pressure. If we convert 8.31 kPa to atm, we get 0.0821 atm, hence both R values are the same and life is good.

9. josephacarmona says:

1. So 27.3 g Fe is 0.4911 mol (27.3 / 55.58). 45.8 g O is 2.862 mol. Because of the stoichiometry, 3/4 as much oxygen as iron gets consumed. For the 0.4911 mol Fe consumed, 0.3683 mol O is consumed. 2.862-0.3863 = 2.4757 mol, or 39.6 g O is left.

2. Hydrogen is the limiting reactant for these quantities. 2/3 as much ammonia is produced as hydrogen is consumed, so for 5.0 mol H2, that's 3.33 mol NH3.

Make sense?