Methane burns in air according to this equation:

Question

C*H_{4}(g) + 2O_{2}(g) -> C*O_{2}(g) + 2H_{2}*O(l)

The table shows the molar masses of substances involved in the reaction.

Molar Masses

Compound
C*H_{4}
C*O_{2}
H_{2}*O
O_{2}
Molar Mass (g / m * ol)
16.0
44.0
18.0
32.0

How many grams of water will be produced if 64.0 g of C*H_{4} burns in excess O_{2}

A 8.0g

B. 36.0 g

C. 72.0 g

D. 144.0 g

Answer 1

To determine how many grams of water (H₂O) will be produced when 64.0 g of methane (CH₄) burns in excess O₂, we first need to follow these steps:

1. Calculate the number of moles of CH₄.

   The molar mass of CH₄ (methane) is given as 16.0 g/mol. \[ \text{Moles of CH₄} = \frac{\text{mass}}{\text{molar mass}} = \frac{64.0 , \text{g}}{16.0 , \text{g/mol}} = 4.0 , \text{mol} \]

2. Use the stoichiometry of the reaction.

   The balanced chemical equation shows that: \[ \text{1 mol CH₄} \rightarrow \text{2 mol H₂O} \] From the reaction, we see that for every 1 mole of CH₄, 2 moles of H₂O are produced. Thus, for 4.0 moles of CH₄: \[ \text{Moles of H₂O} = 4.0 , \text{mol CH₄} \times 2 , \text{mol H₂O/mol CH₄} = 8.0 , \text{mol H₂O} \]

3. Calculate the mass of H₂O produced.

   Now we calculate the mass of water produced using its molar mass (18.0 g/mol): \[ \text{Mass of H₂O} = \text{moles} \times \text{molar mass} = 8.0 , \text{mol} \times 18.0 , \text{g/mol} = 144.0 , \text{g} \]

Thus, the answer is D. 144.0 g of water will be produced when 64.0 g of CH₄ burns in excess O₂.