Chemistry 10th Edition

Balanced equation : $CH_4 + 2O_2 -- \gt CO_2 + 2H_2O$ Mass of $O_2$ that reacts with 44.5 g of methane: 178 g.
1. Write the equation, and balance it: $CH_4 + O_2 -- \gt CO_2 + H_2O$ Begin with the elements that appear only one in each side: - Carbon is already balanced: - Balance the $H$: Since the subscript for hydrogen in the reactants is 4, the coefficient for $H_2O$ will be $\frac{4}{2} = 2$. * Each water molecule has 2 hydrogens. $CH_4 + O_2 -- \gt CO_2 + 2H_2O$ - Now, balance the $O$: There is a total of 4 oxygens on the products side. Therefore, we need to put a $\frac{4}{2} = 2$ as the coefficient of $O_2$. $CH_4 + 2O_2 -- \gt CO_2 + 2H_2O$ --------------------- - Calculate the mass of oxygen that reacts with 44.5 g of methane: 1. Calculate the number of moles of $CH_4$: 12.01* 1 + 1.008* 4 = 16.04g/mol $44.5g \times \frac{1 mol}{ 16.04g} = 2.774mol (CH_4)$ According to the balanced equation: The ratio of $CH_4$ to $O_2$ is 1 to 2: $2.774 mol (CH_4) \times \frac{ 2 mol(O_2)}{ 1 mol (CH_4)} = 5.548mol (O_2)$ 2. Calculate the mass of $O_2$: 16.00* 2 = 32.00g/mol $5.548 mol \times \frac{ 32.00 g}{ 1 mol} = 178 g (O_2)$