Question

When methane (CH4) burns, it reacts with oxygen gas to produce carbon dioxide and water. The unbalanced equation for this reaction iCH4(g)+O2(g)→CO2(g)+H2O(g)This type of reaction is referred to as a complete combustion reaction.A)What mass of carbon dioxide is produced from the complete combustion of 1.10×10−3 g of methane?Express your answer with the appropriate units.B)What mass of water is produced from the complete combustion of 1.10×10−3 g of methane?Express your answer with the appropriate units.C)What mass of oxygen is needed for the complete combustion of 1.10×10−3 g of methane?Express your answer with the appropriate units.

Answer 1

Answer:

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:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$     .....(1)

• For methane:

Given mass of methane = $1.10\times 10^{-3}g$

Molar mass of methane = 16 g/mol

Putting values in equation 1, we get:

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

The chemical equation for the combustion of methane follows:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

• For A:

By Stoichiometry of the reaction:

1 mole of methane produces 1 mole of carbon dioxide

So, $6.875\times 10^{-5}mol$ of methane will produce = $\frac{1}{1}\times 6.875\times 10^{-5}=6.875\times 10^{-5}mol$ 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 = $6.875\times 10^{-5}$ moles

Putting values in equation 1, we get:

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

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, $6.875\times 10^{-5}mol$ of methane will produce = $\frac{2}{1}\times 6.875\times 10^{-5}=1.375\times 10^{-4}mol$ of water

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

Molar mass of water = 18 g/mol

Moles of water = $1.375\times 10^{-4}$ moles

Putting values in equation 1, we get:

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

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, $6.875\times 10^{-5}mol$ of methane will react with = $\frac{2}{1}\times 6.875\times 10^{-5}=1.375\times 10^{-4}mol$ 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 = $1.375\times 10^{-4}$ moles

Putting values in equation 1, we get:

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

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