Answer:
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 = [tex]\frac{mass}{molar mass}[/tex]
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 = [tex]\frac{69.1}{169.9}[/tex] = 0.41moles
2. From the balanced equation;
2 moles of AgNO₃ produced 1 moles of Ag₂CrO₄ ;
0.41 moles of AgNO₃ will produce [tex]\frac{0.41}{2}[/tex] = 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
