# Calculate the amount of water formed when "8 g" of "H"_2" is allowed to react completely with oxygen?

Jun 8, 2018

$\approx \text{70 g H"_2"O}$ will produced from $\text{8 g H"_2}$.

#### Explanation:

Balanced equation

$\text{2H"_2 + "O"_2}$$\rightarrow$$\text{2H"_2"O}$

color(red)(1. Calculate mol $\text{H"_2}$ by dividing its given mass by its molar mass $\left(\text{2 g/mol}\right)$. Do this by multiplying by the inverse of the molar mass.

color(blue)(2. Calculate mol $\text{H"_2"O}$ by multiplying mol $\text{H"_2}$ by the mol ratio between $\text{H"_2"O}$ and $\text{H"_2}$ from the balanced equation, with $\text{H"_2"O}$ in the numerator.

color(green)(3. Calculate mass $\text{H"_2"O}$ by multiplying by the molar mass of $\text{H"_2"O}$ $\left(\text{18 g H"_2"O}\right)$.

color(red)8color(black)cancel(color(red)("g H"_2))xx(color(red)(1color(black)cancel(color(red)("mol H"_2))))/(color(black)cancel(color(red)(2"g H"_2)))xx(color(blue)2color(black)cancel(color(blue)("mol H"_2"O")))/(color(blue)2color(black)cancel(color(blue)("mol H"_2)))xx(color(green)18color(green)("g H"_2"O"))/(color(green)1color(black)cancel(color(green)("mol H"_2"O")))="70 g H"_2"O" (rounded to one significant figure)

Jun 8, 2018

72 g

#### Explanation:

The equation for the reaction of hydrogen gas and oxygen gas to form water is: $2 {H}_{2} \left(g\right) + {O}_{2} \left(g\right) \to 2 {H}_{2} O \left(l\right)$

So for every mole of ${H}_{2}$ you will form a mole of water, assuming you have sufficient oxygen to react with it all.

The atomic weight of hydrogen atoms is 1, but the molar mass of diatomic ${H}_{2}$ is 2 g/mol. Therefore, 8 g of ${H}_{2}$ is 4 moles. Therefore you will form 4 moles of water. The molar mass of water is 18 g/mol so you will form 4 x 18 = 72 g of water.