Based on the given chemical reaction, as 31.2 mL of hydrogen are yielded, one computes its moles via the ideal gas equation under the stated conditions as shown below:

Now, since the relationship between hydrogen and magnesium is 1 to 1, one computes its milligrams by following the shown below proportional factor development:

8.702g of HCl

Explanation:

Using the details from the equation;

4HCl(aq) + MnO2(s) → 2H2O(l) + MnCl2(aq) + Cl2(g)

The molecular mass of HCl is; 4 (1 + 35.5) = 146g

The molecular mass of MnO2 is; (54.9 + 32) = 86.9g

From the equation it means;

86.9g of MnO2 reacts with 146g of HCl

if 1 g of MnO2 reacts with 146 ÷ 86.9 of HCl

∴ 5.18g of MnO2 will react with; 146 ÷ 86.9 x 5.18 = 8.702g of HCl

Note: Approximations were made during the determination of the molecular masses.

5.875

Explanation:

[tex]m_{Mg}=30.8mgMg[/tex]

Explanation:

Hello,

Based on the given chemical reaction, as 31.2 mL of hydrogen are yielded, one computes its moles via the ideal gas equation under the stated conditions as shown below:

[tex]n_{H_2}=\frac{PV}{RT}=\frac{754torr*\frac{1atm}{760torr}*0.0312L}{0.082 \frac{atm*L}{mol*K}*298.15K}=1.27x10^{-3}molH_2[/tex]

Now, since the relationship between hydrogen and magnesium is 1 to 1, one computes its milligrams by following the shown below proportional factor development:

[tex]m_{Mg}=1.27x10^{-3}molH_2*\frac{1molMg}{1molH_2}*\frac{24.305gMg}{1molMg}*\frac{1000mgMg}{1gMg}\\m_{Mg}=30.8mgMg[/tex]

Best regards.

The mass of bromine reacted is 160.6 grams.

Explanation:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)

Given mass of aluminium = 18.1 g

Molar mass of aluminium = 27 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of aluminium}=\frac{18.1g}{27g/mol}=0.670mol[/tex]

The chemical equation for the reaction of aluminium and bromide follows:

[tex]2Al+3Br_2\rightarrow 2AlBr_3[/tex]

By Stoichiometry of the reaction:

2 moles of aluminium reacts with 3 moles of bromine gas

So, 0.670 moles of aluminium will react with = [tex]\frac{3}{2}\times 0.670=1.005mol[/tex] of bromine gas.

Now, calculating the mass of bromine gas, we use equation 1:

Moles of bromine gas = 1.005 moles

Molar mass of bromine gas = 159.81 g/mol

Putting values in equation 1, we get:

[tex]1.005mol=\frac{\text{Mass of bromine}}{159.81g/mol}\\\\\text{Mass of bromine}=(1.005mol\times 159.81g/mol)=160.6g[/tex]

Hence, the mass of bromine reacted is 160.6 grams.

Answer : The mass of [tex]Mg[/tex] required is 30.38 mg.

Explanation :

To calculate the moles of hydrogen gas, we use the equation given by ideal gas :

PV = nRT

where,

P = Pressure of hydrogen gas = 754 torr

V = Volume of the hydrogen gas = 31.2 mL = 0.0312 L

n = number of moles of hydrogen gas = ?

R = Gas constant = [tex]62.364\text{ L.torr }mol^{-1}K^{-1}[/tex]

T = Temperature of hydrogen gas = [tex]25^oC=273+25=298K[/tex]

Putting values in above equation, we get:

[tex]754torr\times 0.0312L=n\times 62.364\text{ L.torr }mol^{-1}K^{-1}\times 298K\\\\n=1.266\times 10^{-3}mole[/tex]

Now we have to calculate the moles of [tex]Mg[/tex].

The balanced chemical reaction is:

[tex]Mg(s)+2HCl(aq)\rightarrow MgCl_2(aq)+H_2(g)[/tex]

From the balanced chemical reaction, we conclude that

As, 1 mole of [tex]H_2[/tex] produced from 1 mole [tex]Mg[/tex]

As, [tex]1.266\times 10^{-3}[/tex] mole of [tex]H_2[/tex] produced from [tex]1.266\times 10^{-3}[/tex] mole [tex]Mg[/tex]

Now we have to calculate the mass of [tex]Mg[/tex].

Molar mass of [tex]Mg[/tex] = 24 g/mol

[tex]\text{Mass of }Mg=\text{Moles of }Mg\times \text{Molar mass of }Mg[/tex]

[tex]\text{Mass of }Mg=1.266\times 10^{-3}mole\times 24g/mole=30.38\times 10^{-3}g=30.38mg[/tex]

conversion used : (1 g = 1000 mg)

Therefore, the mass of [tex]Mg[/tex] required is 30.38 mg

1.1g of copper

Explanation: We begin by writing the balanced reaction equation.

So we have : Cu(s) + 2AgNO3 (aq) > Cu(NO3)2 (aq) + 2Ag (s)

Molar mass of AgNO3 = 107.87 + 14.01 + 3(16.0) = 169.88g/mol, for 2 moles we then have 2 x 169.88 =339.76g.

Atomic mass of Copper = 64g approximately

From the equation the following deductions can be made:

339.76g of AgNO3 reacts with 64g of copper

5.65g of AgNO3 would react with 64/ 339.76 x 5.65 =1.062 approx 1.1g of copper.

Answer : 70.906 grams of chlorine reacted with hydrogen

Explanation :

Step 1 : Write balanced chemical equation.

The balanced chemical equation for the reaction between hydrogen and chlorine gas is given below.

[tex]H_{2} (g) + Cl_{2} (g) \rightarrow 2HCl(g)[/tex]

Step 2 : Find moles of H₂ gas.

The moles of H₂ can be found as

[tex]mole = \frac{grams}{MolarMass}[/tex]

We have 2.0200 g of H₂ and molar mass of H₂ is 2.02 g/mol.

Let us plug in these values to find moles of H₂.

[tex]mole =\frac{2.0200g}{2.02g/mol} = 1 mol[/tex]

We have 1 mol of H₂.

Step 3 : Find moles of Cl₂ using mole ratio.

The mole ratio of H₂ and Cl₂ is 1 : 1.

The moles of Cl₂ can be calculated as

[tex]1 mol H_{2} \times\frac{1 mol Cl_{2}}{1 mol H_{2}} = 1 mol Cl_{2}[/tex]

Step 4 : Find grams of Cl₂.

Molar mass of Cl₂ gas is 70.096 g/mol

Mass of Cl₂ = [tex]1 mol Cl_{2} \times\frac{70.906g}{mol} = 70.906 g[/tex]

We have 70.906 grams of Cl₂

3.89 g Mg

Explanation:

2M = 2 mol/L

120 mL = 0.120 L

2 mol/L *0.120 L = 0.240 mol H3PO4

3Mg + 2H3PO4 > Mg3(PO4)2 + 3H2

from reaction 3 mol 2 mol

given 0.240mol x mol

x = 0.240*2/3 = 0.16 mol Mg

M(Mg) = 24.3 g/mol

24.3 g/mol *0.16 mol = 3.89 g Mg

a) Na2SiO3(s) + 8HF(aq) → H2SiF6(aq) + 2NaF(aq) + 3H2O (l)

b) 2.40 moles HF

c) 5.25 grams NaF

d)0.609 grams Na2SiO3

e) 0.89 grams Na2SiO3

Explanation:

Step 1: Data given

Step 2: The balanced equation

Na2SiO3(s) + 8HF(aq) → H2SiF6(aq) + 2NaF(aq) + 3H2O (l)

b) How many moles of HF are needed to react with 0.300 mol of Na2SiO3?

For 1 mol Na2SiO3 we need 8 moles HF to produce 1 mol H2SiF6, 2 moles NaF and 3 moles H2O

For 0.300 moles Na2SiO3 we'll need 8*0.300 = 2.40 moles HF

c. How many grams of NaF form when 0.500 mol of HF reacts with excess Na2SiO3?

For 1 mol Na2SiO3 we need 8 moles HF to produce 1 mol H2SiF6, 2 moles NaF and 3 moles H2O

For 0.500 moles we'll have 0.500 / 4 = 0.125 moles NaF

Mass NaF = 0.125 moles * 41.99 g/mol

Mass NaF = 5.25 grams NaF

d. How many grams of Na2SiO3 can react with 0.800 g of HF?

Moles HF = 0.800 grams / 20.01 g/mol

Moles HF = 0.0399 moles

For 1 mol Na2SiO3 we need 8 moles HF to produce 1 mol H2SiF6, 2 moles NaF and 3 moles H2O

For 8 moles we need 1 moles Na2SiO3 to react

For 0.0399 moles we need 0.0399/8 = 0.00499 moles Na2SiO3

Mass Na2SiO3 = 0.00499 moles * 122.06 g/mol

Mass Na2SiO3 = 0.609 grams Na2SiO3

Using your answer to part d-If you started with 1.5 g of Na2SiO3how many grams would be left after the reaction is complete?

Number of moles HF = 0.0399 moles

Number of moles Na2SiO3 = 1.5 grams / 122.06 g/mol = 0.0123 moles

For 1 mol Na2SiO3 we need 8 moles HF to produce 1 mol H2SiF6, 2 moles NaF and 3 moles H2O

For 8 moles we need 1 moles Na2SiO3 to react

For 0.0399 moles we need 0.0399/8 = 0.00499 moles Na2SiO3

There will remain 0.0123 - 0.00499 = 0.00731 moles Na2SiO3

Mass Na2SiO3 remaining = 0.00731 * 122.06 g/mol = 0.89 grams Na2SiO3

30.7 mg of Magnesium

Explanation: The equation of reaction is;

Mg(s) + 2HCl(aq) > MgCl2(aq) + H2(g).

I mol of Magnessium produces 1 mol of Hydrogen gas.

Pressure,p= 754 torr, volume, v= 31.2 ml( which is also equals to 0.0312L) , R= 62.36367L torr/kmol, temperature, T = 25 + 273.15)

Step 1: find the number of moles using PV = nRT.

Making n(number of mole) the subject of the formula, we have;

n = PV/RT

Substitute the above parameters into the formula, we have;

n= (754 Torr) x (0.0312 L) / ((62.36367 L torr/K mol) x (25.0 + 273.15) K) = 0.001265 mol of Hydrogen gas.

Step 2: using n = mass/molar mass.

Mass = n × molar mass

(0.001265 mol of H2) x (1 mol Mg / 1 mol H2) x (24.30506 g Mg/mol) = 0.030745 g = 30.7 mg of Magnessium