# How many grams of silver chromate are produced when 250. mL of 1.500 M Na2CrO4 are added to excess silver nitrate? Please balance the equation before solving the problem.

May 31, 2018

The mass of ${\text{Ag"_2"CrO}}_{4}$ produced is 124 g.

#### Explanation:

There are four steps involved in this stoichiometry problem:

1. Write the balanced chemical equation
2. Calculate the moles of ${\text{Na"_2"CrO}}_{4}$
3. Use the molar ratio from the balanced equation to calculate the moles of ${\text{Ag"_2"CrO}}_{4}$
4. Calculate the mass of ${\text{Ag"_2"CrO}}_{4}$

Step 1. Write the balanced chemical equation.

${M}_{r} : \textcolor{w h i t e}{m m m m m m m m m m m m m l l} 331.73$
$\textcolor{w h i t e}{m m m} {\text{2AgNO"_3 + "Na"_2"CrO"_4 →"Ag"_2"CrO"_4 + "2NaNO}}_{3}$

Step 2. Calculate the moles of ${\text{Na"_2"CrO}}_{4}$

${\text{Moles of Na"_2"CrO"_4 = 0.250 color(red)(cancel(color(black)("L Na"_2"CrO"_4))) × ("1.500 mol Na"_2"CrO"_4)/(1 color(red)(cancel(color(black)("L Na"_2"CrO"_4)))) = "0.3750 mol Na"_2"CrO}}_{4}$

Step 3. Calculate the moles of ${\text{Ag"_2"CrO}}_{4}$

The molar ratio from the balanced equation is

${\text{Ag"_2"CrO"_4:"Na"_2"CrO}}_{4} = 1 : 1$.

${\text{Moles of Ag"_2"CrO"_4 = 03750 color(red)(cancel(color(black)("mol Na"_2"CrO"_4))) × ("1 mol Ag"_2"CrO"_4)/(1 color(red)(cancel(color(black)("mol Na"_2"CrO"_4)))) = "0.3750 mol Ag"_2"CrO}}_{4}$

Step 4. Calculate the mass of ${\text{Ag"_2"CrO}}_{4}$

${\text{Mass of Ag"_2"CrO"_4 = 0.3750 color(red)(cancel(color(black)("mol Ag"_2"CrO"_4))) × ("331.73 g Ag"_2"CrO"_4)/(1color(red)(cancel(color(black)("mol Ag"_2"CrO"_4)))) = "124 g Ag"_2"CrO}}_{4}$