Question

methane burns in air according to this equation: ch₄(g) + 2o₂(g) → co₂(g) + 2h₂o(l)
the table shows the molar masses of substances involved in the reaction.
molar masses

compound	molar mass (g/mol)
co₂	44.0
h₂o	18.0
o₂	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

Explanation:

Step1: Calculate moles of CH₄

$n_{CH_4}=\frac{m_{CH_4}}{M_{CH_4}}=\frac{64.0\ g}{16.0\ g/mol}=4.0\ mol$

Step2: Determine mole - ratio of CH₄ to H₂O

From the balanced equation $CH_4(g)+2O_2(g)
ightarrow CO_2(g) + 2H_2O(l)$, the mole - ratio of $CH_4$ to $H_2O$ is $1:2$. So, if $n_{CH_4} = 4.0\ mol$, then $n_{H_2O}=2\times n_{CH_4}=2\times4.0\ mol = 8.0\ mol$

Step3: Calculate mass of H₂O

$m_{H_2O}=n_{H_2O}\times M_{H_2O}=8.0\ mol\times18.0\ g/mol = 144.0\ g$

Answer:

D. 144.0 g