# Calcium hydride reacts with water to form calcium hydroxide and hydrogen gas. The balanced equation to this is CaH_2+2H_2O----Ca(OH)_2+2H_2 How many grams of calcium hydride are needed to form 8.400 g of hydrogen?

Apr 1, 2016

You need ${\text{87.69 g of CaH}}_{2}$ to form ${\text{8.400 g of H}}_{2}$.

#### Explanation:

There are four steps to answering this type of stoichiometry problem.

1. Write the balanced equation for the reaction.
2. Use the molar mass of ${\text{H}}_{2}$ to convert grams of ${\text{H}}_{2}$ to moles of ${\text{H}}_{2}$.
3. Use the molar ratio from the balanced equation to convert moles of ${\text{H}}_{2}$ to moles of ${\text{CaH}}_{2}$.
4. Use the molar mass of ${\text{CaH}}_{2}$ to convert moles of ${\text{CaH}}_{2}$ to grams of ${\text{CaH}}_{2}$.

Step 1. Write the balanced equation.

${\text{CaH"_2 + "2H"_2"O" → "Ca(OH)"_2 + "2H}}_{2}$

Step 2. Convert grams of ${\text{H}}_{2}$ to moles of ${\text{H}}_{2}$.

The molar mass of ${\text{H}}_{2}$ is 2.016 g/mol.

8.400 cancel("g H"_2) × ("1 mol H"2)/(2.016 cancel("g H"2)) = "4.1667 mol H"_2

Step 3. Use the molar ratio to calculate the moles of ${\text{CaH}}_{2}$.

From the balanced equation, the molar ratio is ${\text{1 mol CaH"_2:"2 mol H}}_{2}$.

4.1667 cancel("mol H"_2) × ("1 mol CaH"_2)/(2 cancel("mol H"_2)) = "2.0833 mol CaH"_2

Step 4. Convert moles of ${\text{CaH}}_{2}$ to grams of ${\text{CaH}}_{2}$.

The molar mass of ${\text{CaH}}_{2}$ is 42.09 g/mol.

2.0833 cancel("mol CaH"_2) × ("42.09 g CaH"_2)/(1 cancel("mol CaH"_2)) = "87.69 g CaH"2

The reaction requires ${\text{87.69 g CaH}}_{2}$.

This video link allows you to furthur practice similar problems.