# Question e15e7

May 1, 2017

You must use $\text{69.68 g of Na"_2"S}$ to produce ${\text{50.00 g of TiS}}_{2}$.

#### 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 ${\text{TiS}}_{2}$ to convert grams of ${\text{TiS}}_{2}$ to moles of ${\text{TiS}}_{2}$.
3. Use the molar ratio from the balanced equation to convert moles of ${\text{TiS}}_{2}$ to moles of $\text{Na"_2"S}$.
4. Use the molar mass of $\text{Na"_2"S}$ to convert grams of $\text{Na"_2"S}$ to moles of $\text{Na"_2"S}$.

Step 1. Write the balanced equation.

${\text{2Na"_2"S" + "TiCl"_4 → "4NaCl" + "TiS}}_{2}$

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

The molar mass of ${\text{TiS}}_{2}$ is 112.00 g/mol.

50.00 cancel("g TiSO"_2) × ("1 mol TiSO"_2)/(112.00 cancel("g TiS"_2)) = "0.446 43 mol TiS"_2#

Step 3. Use the molar ratio to calculate the moles of $\text{Na"_2"S}$.

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

$\text{0.446 43" cancel("mol TiS"_2) × ("2 mol Na"_2"S")/(1 cancel("mol TiS"_2)) = "0.892 86 mol Na"_2"S}$

Step 4. Convert moles of $\text{Na"_2"S}$ to grams of $\text{Na"_2"S}$.

$\text{0.892 86" color(red)(cancel(color(black)("mol Na"_2"S"))) × ("78.04 g Na"_2"S")/(1 color(red)(cancel(color(black)("mol Na"_2"S")))) = "69.68 g Na"_2"S}$

You need to use $\text{69.68 g of Na"_2"S}$.