# Question 49071

May 28, 2017

#### Answer:

The reaction would produce 2.01 × 10^6 color(white)(l)"g H"_2"O"_2.

#### Explanation:

Step 1. Write the balanced chemical equation

M_text(r):color(white)(mmmmmm) 34.01"#
$\textcolor{w h i t e}{m m} {\text{H"_2 + "O"_2 → "H"_2"O}}_{2}$

Step 2. Calculate the moles of ${\text{O}}_{2}$

${\text{Moles of O"_2 = 3.56 × 10^28 color(red)(cancel(color(black)("molecules O"_2))) × "1 mol O"_2/(6.022 × 10^23 color(red)(cancel(color(black)("molecules O"_2)))) = "59 120 mol O}}_{2}$

Step 3. Calculate the moles of ${\text{H"_2"O}}_{2}$

${\text{Moles of H"_2"O"_2 = 59 120 color(red)(cancel(color(black)("mol O"_2))) × ("1 mol H"_2"O"_2)/(1 color(red)(cancel(color(black)("mol O"_2)))) = "59 120 mol H"_2"O}}_{2}$

Step 4. Calculate the mass of ${\text{H"_2"O}}_{2}$

${\text{Mass of H"_2"O"_2 = "59 120" color(red)(cancel(color(black)("mol H"_2"O"_2))) × ("34.01 g H"_2"O"_2)/(1 color(red)(cancel(color(black)("mol H"_2"O"_2)))) = 2.01 × 10^6 color(white)(l)"g H"_2"O}}_{2}$