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

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