For the reaction 2AgNO3+Na2CrO4⟶Ag2CrO4+2NaNO3

how many grams of silver chromate, Ag2CrO4, are produced from 69.1 g of silver nitrate, AgNO3?

66.4g

Explanation:

Given parameters:

      Mass of AgNO₃  = 69.1g

Unknown:

    Mass of Ag₂CrO₄

Solution:

  2AgNO₃+ Na₂CrO₄   ⟶  Ag₂CrO₄  +  2NaNO₃

The given equation is balanced.

 To solve this problem, solve from the known to the unknown;

1. Find the number of moles of AgNO₃;

    Number of moles =

atomic mass of Ag = 107.9g/mol

                           N  = 14g/mol

                           O = 16g/mol  

  Molar mass of AgNO₃ = 107.9 + 14 + 3(16) = 169.9g/mol

 Number of moles =  = 0.41moles

2. From the balanced equation;

2 moles of AgNO₃ produced 1 moles of Ag₂CrO₄ ;

0.41 moles of AgNO₃ will produce     = 0.2moles of Ag₂CrO₄

3:

      Mass of Ag₂CrO₄ = number of moles x molar mass

    Atomic mass of Ag = 107.9g/mol

                               Cr = 52g/mol

                                0 = 16g/mol

     Molar mass  = 2(107.9) + 52 + 4(16) = 331.8g/mol

  so, mass of Ag₂CrO₄ = 0.2 x 331.8 = 66.4g

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