What mass of carbon dioxide is produced from the complete combustion of 8.00×10−3 g of methane? express your answer with the appropriate units.

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What mass of carbon dioxide is produced from the complete combustion of 8.00×10−3 g of methane? express your answer with the appropriate units.

1.406 mass of carbon dioxide is produced

- [tex]1.76*10^-^2[/tex] g of carbon dioxide.

Solution:- The balanced equation for the combustion of methane is:

[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]

There is 1:1 mol ratio between methane and carbon dioxide. Grams of methane are converted to moles and then using mol ratio we get the moles of carbon dioxide that could further be converted to grams. The calculations are shown as:

[tex]6.40*10^-^3gCH_4(\frac{1molCH_4}{16gCH_4})(\frac{1molCO_2}{1molCH_4})(\frac{44gCO_2}{1molCO_2})[/tex]

= [tex]1.76*10^-^2gCO_2[/tex]

So, complete combustion of given amount of methane gives [tex]1.76*10^-^2gCO_2[/tex] .

17.6×10^-3g of CO2

Explanation:

We first look at the stoichiometry of the balanced reaction equation. One mole of methane produces one mole of carbon dioxide. Hence 16g of methane yields 44g of carbon dioxide. If we now composed this with the given 6.40×10^-3g of methane as shown in the solution attached, we obtain the answer stated above.

[tex]What mass of carbon dioxide is produced from the complete combustion of 6.40×10−3 gg of methane? ch[/tex]

CH₄ + 2O₂ → CO₂ + 2H₂O

From the equation, we know that methane and carbon dioxide have the same number of moles.

[tex]no. of moles = \frac{mass}{molar mass}[/tex]

no. of moles of CO₂ produced = no. of moles of methane

= 4.5 × 10⁻³ ÷ (12 + 1×4)

= 2.8125 × 10⁻⁴

∴ mass of CO₂ = 2.8125 × 10⁻⁴ × (12 + 16×2)

= 12.375 × 10⁻³ g

0.0159 g of carbon dioxide

0.0195g of CO2 is produced

Explanation:

CH4 + 2O2 —> CO2 + 2H2O

Molar Mass of CO2 = 12 + (16x2) = 12 + 32 = 44g/mol

Molar Mass of CH4 = 12 + (4x1) = 12 + 4 = 16g/mol

From the equation,

16g of CH4 produce 44g of CO2.

Therefore 7.10×10^−3 of CH4 will produce Xg of CO2 i.e

Xg of CO2 = (7.10×10^−3x44) /16 = 0.0195g

For A: The mass of carbon dioxide produced is 0.00303 grams.

For B: The mass of water produced is 0.0025 grams.

For C: The mass of oxygen gas produced is 0.0044 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)

For methane:

Given mass of methane = [tex]1.10\times 10^{-3}g[/tex]

Molar mass of methane = 16 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of methane}=\frac{1.10\times 10^{-3}g}{16g/mol}=6.875\times 10^{-5}mol[/tex]

The chemical equation for the combustion of methane follows:

[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]

For A:

By Stoichiometry of the reaction:

1 mole of methane produces 1 mole of carbon dioxide

So, [tex]6.875\times 10^{-5}mol[/tex] of methane will produce = [tex]\frac{1}{1}\times 6.875\times 10^{-5}=6.875\times 10^{-5}mol[/tex] of carbon dioxide

Now, calculating the mass of carbon dioxide from equation 1, we get:

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = [tex]6.875\times 10^{-5}[/tex] moles

Putting values in equation 1, we get:

[tex]6.875\times 10^{-5}mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(6.875\times 10^{-5}mol\times 44g/mol)=0.00303g[/tex]

Hence, the mass of carbon dioxide produced is 0.00303 grams.

For B:

By Stoichiometry of the reaction:

1 mole of methane produces 2 moles of water

So, [tex]6.875\times 10^{-5}mol[/tex] of methane will produce = [tex]\frac{2}{1}\times 6.875\times 10^{-5}=1.375\times 10^{-4}mol[/tex] of water

Now, calculating the mass of water from equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = [tex]1.375\times 10^{-4}[/tex] moles

Putting values in equation 1, we get:

[tex]1.375\times 10^{-4}mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(1.375\times 10^{-4}mol\times 18g/mol)=0.0025g[/tex]

Hence, the mass of water produced is 0.0025 grams.

For C:

By Stoichiometry of the reaction:

1 mole of methane reacts with 2 moles of oxygen gas

So, [tex]6.875\times 10^{-5}mol[/tex] of methane will react with = [tex]\frac{2}{1}\times 6.875\times 10^{-5}=1.375\times 10^{-4}mol[/tex] of oxygen gas

Now, calculating the mass of oxygen gas from equation 1, we get:

Molar mass of oxygen gas = 32 g/mol

Moles of oxygen gas = [tex]1.375\times 10^{-4}[/tex] moles

Putting values in equation 1, we get:

[tex]1.375\times 10^{-4}mol=\frac{\text{Mass of oxygen gas}}{32g/mol}\\\\\text{Mass of oxygen gas}=(1.375\times 10^{-4}mol\times 32g/mol)=0.0044g[/tex]

Hence, the mass of oxygen gas produced is 0.0044 grams.

22×10−3 g

Explanation:

Combustion equation of methane:

CH4 + 2O2 → CO2 + 2H2O

Given data:

mass of methane= 8.00 ×10−3 g

mass of carbon dioxide= ?

Solution 1:

m (CH4) = 8.00 ×10−3 g

v=m /M

v(CH4) = m (CH4)/ M (CH4)

v(CH4)= 8.00 ×10−3 g / 16 g/ mol

v(CH4)= 0.5 ×10−3 mol

according to balance chemical equation:

v(CH4) : v(CO2) = 1:1

v(CH4) = v(CO2)

M(CO2) = 44 g/mol

m (CO2) = M(CO2) × v(CO2)

m (CO2) = 44 g/mol ×0.5 ×10−3 mol = 22×10−3 g

Second method:

molecular weight of methane = 16 g/mol

molecular weight of carbon dioxide = 44 g/mol

mass of methane= 8.00×10−3 g

mass of carbon dioxide= ?

Solution:

mass of carbon dioxide= mass of methane × molecular weight of carbon dioxide / molecular weight of methane

mass of carbon dioxide= 8.00×10−3 g× 44 g/mol / 16 g/mol

mass of carbon dioxide= 22 × 10−3 g

mCO2= 49.6932 kgCO2

Explanation:

Hello! Let's solve this!

First we propose the balanced equation C3H8 + 5O2 ---> 3CO2 + 4H2O

We see that each mole of C3H8 (propane) we get 3 moles of CO2

From the propane volume we can obtain the grams of propane used.

molpropane = 26.5L * (1000mL / 1L) * (0.621g / 1mL) * (1mol / 44g) = 374.01mol propane

mCO2 = 374.01molC3H8 * (3molCO2 / 1molC3H8) * (44gCO2 / 1molCO2) = 49369.32g * (1kg / 1000g) = 49.6932 kgCO2

mCO2= 49.6932 kgCO2

The answer for the following problem is explained below.

Therefore 16.5 × 10^-3 grams of carbon dioxide is produced from the complete combustion.

Explanation:

Given:

mass of methane = 6.00 × 10^-3 grams

[tex]CH_{4}[/tex] + [tex]O_{2}[/tex] → [tex]CO_{2}[/tex] + [tex]H_{2}O[/tex]

Firstly balance the following equation:

Before balancing the equation:

[tex]CH_{4} + O_{2}[/tex] → [tex]CO_{2} + H_{2} O[/tex]

After balancing the equation:

[tex]CH_{4} + 2O_{2}[/tex] → [tex]CO_{2} + 2 H_{2} O[/tex]

where;

[tex]CH_{4}[/tex] represents methane molecule

[tex]O_{2}[/tex] represents oxygen molecule

[tex]CO_{2}[/tex] represents carbon dioxide molecule

[tex]H_{2}O[/tex] represents water molecule

[tex]CH_{4}[/tex] +2 [tex]O_{2}[/tex] → [tex]CO_{2}[/tex] + 2[tex]H_{2}O[/tex]

16 grams of methane → 44 grams of carbon dioxide

6 × 10^-3 grams of methane → ?

= [tex]\frac{44*6.00*10^{-3} }{16}[/tex]

= 16.5 × 10^-3 grams of carbon dioxide is produced from the complete combustion.

Therefore 16.5 × 10^-3 grams of carbon dioxide is produced from the complete combustion.