Question

consider combustion of natural gas (methane) in oxygen to from water and carbon dioxide: ch4 + o2 --> h2o + co2. balance the equation, and find the number of grams of water produced by the combustion of 15.0 g of methane. 33.7 15.0 16.8 8.4

Explanation:

Step1: Balance the chemical equation

The balanced equation is $CH_4 + 2O_2
ightarrow 2H_2O+CO_2$.

Step2: Calculate moles of methane

The molar - mass of $CH_4$ is $M_{CH_4}=(12.01 + 4\times1.01)\ g/mol=16.05\ g/mol$. The number of moles of $CH_4$, $n_{CH_4}=\frac{m_{CH_4}}{M_{CH_4}}=\frac{15.0\ g}{16.05\ g/mol}\approx0.935\ mol$.

Step3: Determine moles of water based on stoichiometry

From the balanced equation, the mole - ratio of $CH_4$ to $H_2O$ is $1:2$. So, $n_{H_2O}=2\times n_{CH_4}=2\times0.935\ mol = 1.87\ mol$.

Step4: Calculate mass of water

The molar - mass of $H_2O$ is $M_{H_2O}=(2\times1.01 + 16.00)\ g/mol = 18.02\ g/mol$. The mass of $H_2O$, $m_{H_2O}=n_{H_2O}\times M_{H_2O}=1.87\ mol\times18.02\ g/mol\approx33.7\ g$.

Answer:

33.7