# Question da554

Mar 6, 2017

$\text{0.306 g H"_2"O}$

#### Explanation:

Start by writing the balanced chemical equation that describes the synthesis of water

$\textcolor{b l u e}{2} {\text{H"_ (2(g)) + "O"_ (2(g)) -> color(purple)(2)"H"_ 2"O}}_{\left(l\right)}$

Notice that the reaction consumes $\textcolor{b l u e}{2}$ moles of hydrogen gas for every $1$ mole of oxygen gas that takes part in the reaction and produces $\textcolor{p u r p \le}{2}$ moles of water.

You already have the number of moles of oxygen gas present in the container, so use the molar mass of hydrogen gas to convert the number of grams of this reactant to moles

0.0343 color(red)(cancel(color(black)("g"))) * "1 mole H"_2/(2.016color(red)(cancel(color(black)("g")))) = "0.0170 moles H"_2

Now, how many moles of oxygen gas would be needed to ensure that all the moles of hydrogen gas react?

0.0170 color(red)(cancel(color(black)("moles H"_2))) * "1 mole O"_2/(color(blue)(2)color(red)(cancel(color(black)("moles H"_2)))) = "0.00850 moles O"_2

Since you have more moles of oxygen gas than required, you can say that the oxygen gas will be in excess. In other words, hydrogen gas will be the limiting reagent because it will be completely consumed before all the moles of oxygen are consumed by the reaction.

So, the reaction will consume $0.0170$ moles of hydrogen gas and produce

0.0170 color(red)(cancel(color(black)("moles O"_2))) * (color(purple)(2)color(white)(.)"moles H"_2"O")/(color(blue)(2)color(red)(cancel(color(black)("moles H"_2)))) = "0.0170 moles H"_2"O"#

To convert this to grams, use the molar mass of water

$0.0170 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles H"_2"O"))) * "18.015 g"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = color(darkgreen)(ul(color(black)("0.306 g}}}}$

The answer is rounded to three sig figs.