# Question c71a5

Jun 16, 2015

Mass of oxygen required: $\text{16 g}$
Mass of water produced: $\text{18 g}$

#### Explanation:

The first thing you need to do is write the balanced chemical equation for this synthesis reaction

$\textcolor{red}{2} {H}_{2 \left(g\right)} + {O}_{2 \left(g\right)} \to 2 {H}_{2} {O}_{\left(l\right)}$

Notice that you have a $\textcolor{red}{2} : 1$ mole ratio between hydrogen gas and oxygen gas. This tells you that, regardless of how many moles of hydrogen react, you'll always need 2 times fewer moles of oxygen in order for the reaction to take place.

Use hydrogen's molar mass to determine how many moles of hydrogen are present in your 2-g sample

2cancel("g") * ("1 mole "H_2)/(2.02 cancel("g")) = "0.99 moles" ${H}_{2}$

This many moles of hydrogen would need

0.99cancel("moles"H_2) * ("1 mole "O_2)/(color(red)(2)cancel("moles"H_2)) = "0.495 moles" ${O}_{2}$

At the same time, this many moles of hydrogen would produce

0.99cancel("moles"H_2) * ("2 moles "H_2O)/(color(red)(2)cancel("moles"H_2)) = "0.99 moles" ${H}_{2} O$

Use the molar masses of oxygen and water, respectively, to determine how many grams of each the reaction will need/produce.

0.495cancel("moles") * "32.0 g"/(1cancel("mole"O_2)) = "15.8 g" ${O}_{2}$

and

0.99cancel("moles"H_2O) * "18.02 g"/(1cancel("mole"H_2O)) = "17.8 g"# ${H}_{2} O$

Since you only gave one sig fig for the mass of hydrogen, the answers must be rounded to one sig fig. However, I'll leave them rounded to two sig figs

${m}_{{O}_{2}} = \textcolor{g r e e n}{\text{16 g}}$

${m}_{{H}_{2} O} = \textcolor{g r e e n}{\text{18 g}}$