Methane burns in air according to this equation:

Question

CH4(9) +202(9) CO2(g) + 2 H2O(1)
The table shows the molar masses of substances involved in the reaction.
Molar Masses
Compound
Molar Mass (g/mol)
CH4
16.0
CO2
44.0
H2O
18.0
02
32.0
How many grams of water will be produced if 64.0 g of CH4 burns in excess O2?
A. 8.0 g
B. 36.0 g
c. 72.0 g
D. 144.0 g

Answer 1

To determine how many grams of water (\(H_2O\)) will be produced from the combustion of 64.0 g of methane (\(CH_4\)), we will follow these steps:

1. Calculate the number of moles of \(CH_4\): \[ \text{Moles of } CH_4 = \frac{\text{mass}}{\text{molar mass}} = \frac{64.0 , \text{g}}{16.0 , \text{g/mol}} = 4.0 , \text{mol} \]

2. Use the stoichiometry of the reaction: The balanced equation is: \[ CH_4 + 2 O_2 \rightarrow CO_2 + 2 H_2O \] From the equation, we see that 1 mole of \(CH_4\) produces 2 moles of \(H_2O\). Therefore, if we have 4 moles of \(CH_4\): \[ \text{Moles of } H_2O = 4.0 , \text{mol} \times 2 = 8.0 , \text{mol} \]

3. Calculate the mass of water produced: \[ \text{Mass of } H_2O = \text{moles} \times \text{molar mass} = 8.0 , \text{mol} \times 18.0 , \text{g/mol} = 144.0 , \text{g} \]

Thus, the answer is \(D\) 144.0 g of water will be produced.