For the reaction

2

AgNO

3

+

Na

2

CrO

4

⟶

Ag

2

CrO

4

+2

NaNO

3

2AgNO3+Na2CrO4⟶Ag2CrO4+2NaNO3

how many grams of sodium chromate, Na2CrO4, are needed to react completely with

12.1

12.1

g of silver nitrate, AgNO3?

Mass of Na₂CrO₄ = 5.75 g

Explanation:

First of all we will write the balance chemical equation.

2AgNO₃ + Na₂CrO₄ → Ag₂CrO₄ + 2NaNO₃

Now we will calculate the moles of AgNO₃.

Number of moles = mass / molar mass

Molar mass of AgNO₃ = 107.87 + 14 + 3 * 16 = 169.87 g/mol

Number of moles = mass / molar mass

Number of moles = 12.1 g / 169.87 g/mol = 0.071 mol

Now we will compare the moles of AgNO₃ and Na₂CrO₄ from balance chemical equation.

AgNO₃ : Na₂CrO₄

2 : 1

0.071 : 1/2 * 0.071 = 0.0355

Now we will calculate the mass of Na₂CrO₄.

Molar mass of Na₂CrO₄ = 23*2 + 52 + 16*4 = 162 g/mol

Mass of Na₂CrO₄ = number of moles * molar mass

Mass of Na₂CrO₄ = 0.0355 mol * 162 g/mol

Mass of Na₂CrO₄ = 5.75 g