How many grams of $text{H}_2text{O}$ are produced by the complete combustion of 16g of $text{CH}_4$?

💡 Explanation:

The balanced chemical equation for the complete combustion of methane is $text{CH}_4 + 2text{O}_2 rightarrow text{CO}_2 + 2text{H}_2text{O}$. The molar mass of methane ($text{CH}_4$) is approximately $16 text{ g/mol}$ and the molar mass of water ($text{H}_2text{O}$) is $18 text{ g/mol}$. Step 1: Convert mass of $text{CH}_4$ to moles. $text{Moles of } text{CH}_4 = text{Mass} / text{Molar Mass} = 16text{g} / 16 text{ g/mol} = 1 text{ mole}$. Step 2: Use stoichiometry to find moles of $text{H}_2text{O}$. The equation shows a $1:2$ mole ratio between $text{CH}_4$ and $text{H}_2text{O}$. $text{Moles of } text{H}_2text{O} text{ produced} = 1 text{ mole } text{CH}_4 times (2 text{ moles } text{H}_2text{O} / 1 text{ mole } text{CH}_4) = 2 text{ moles}$. Step 3: Convert moles of $text{H}_2text{O}$ to mass. $text{Mass of } text{H}_2text{O} = text{Moles} times text{Molar Mass} = 2 text{ moles} times 18 text{ g/mol} = 36 text{ grams}$.