# Question 2f29b

May 24, 2017

The reaction produces 10.5 g of water.

#### Explanation:

Step 1. Write the balanced chemical equation

${M}_{\textrm{r}} : \textcolor{w h i t e}{m m m m m m l} 18.02$
$\textcolor{w h i t e}{m m} \text{2H"_2 + "O"_2 → "2H"_2"O}$

Step 2. Calculate the moles of hydrogen

We can use the Ideal Gas Law to calculate the number of moles.

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} p V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

We can rearrange this formula to get

$n = \frac{p V}{R T}$

In this problem,

p = 801 color(red)(cancel(color(black)("mmHg"))) × "1 atm"/(760 color(red)(cancel(color(black)("mmHg")))) = "1.054 atm"
$V = \text{13.74 L}$
$R = \text{0.082 06 L·atm·K"^"-1""mol"^"-1}$
$T = \text{(30.0 + 273.15) K" = "303.15 K}$

n = (1.054 cancel("atm") × 13.74 cancel("L"))/("0.082 06" cancel("L·atm·K"^"-1")"mol"^"-1" × 303.15 cancel("K")) = "0.5821 mol"#

Step 3. Calculate the moles of water formed

$\text{Moles of H"_2"O" = 0.5821 color(red)(cancel(color(black)("mol H"_2))) × ("2 mol H"_2"O")/(2 color(red)(cancel(color(black)("mol H"_2)))) = "0.5821 mol H"_2"O}$

Step 4. Calculate the mass of water formed

$\text{Mass of H"_2"O" = 0.5821 cancel("mol H"_2"O") × ("18.02 g H"_2"O")/(1 cancel("mol H"_2"O")) = "10.5 g H"_2"O}$