Answer to Question #195282 in General Chemistry for Jerick

Question

How many grams of Ag₂ Cr0₄ will precipitate when 86ml of 0.2 molar Ag₂Cr0₄ is added to 500ml

of 6.3 molar K₂CrO₄? Molar mass of Ag₂CrO4 = 331.74 g/mol

Balanced Equation: 2AgNO3 + K2CrO₄ Ag₂ Cr0₄ + 2KN0₃

Answer 1

Moles of AgNO₃

molarity = moles of solute/liters of solution = mol/L

moles of solute = liters of solution × molarity

Convert mL to L.

86.0 mL × 1 L/1000 mL = 0.0860 L

moles AgNO₃ = 0.0860 L × 0.2 mol AgNO₃ = 0.0172 mol AgNO₃

Moles of K₂CrO4

Convert mL to L.

50.0 mL × 1 L/1000 mL = 0.0500 L

moles K₂CrO4 = 0.3 mol K₂CrO4 × 0.0500 L = 0.0150 mol K₂CrO4

Use stoichiometry to calculate the moles of Ag₂CrO4 that can be produced by each reactant, using the mole ratios between each reactant and Ag₂CrO4 in the balanced equation.

Mol AgNO₃ → Mol Ag₂CrO4

0.0172 mol AgNO₃ × 1 mol Ag₂CrO4/2 mol AgNO₃ = 0.00860 mol Ag₂CrO4

Mol K₂CrO4 → Mol Ag₂CrO4

0.0150 mol K₂CrO4 × 1 mol Ag₂CrO4/1 mol K₂CrO4 = 0.0150 mol Ag₂CrO4

Moles of Ag₂CrO4 produced by each reactant.

AgNO₃ → 0.00860 mol Ag₂CrO4

K₂CrO4 → 0.0150 mol Ag₂CrO4

KNO₃ produces the least number of moles of Ag₂CrO4, so it is the limiting reagent, and determines the amount of Ag₂CrO4 that can be produced.

Calculate the mass of Ag₂CrO4 in 0.00860 mol.

mass (m) = moles (n)/molar mass (M)

n(Ag₂CrO4) = 0.00860 mol Ag₂CrO4

M(Ag₂CrO4) = (2 × 107.868 g/mol Ag) + 51.996 g/mol Cr) + (4 × 15.999 g/mol O) = 331.728 g Ag₂CrO4/mol Ag₂CrO4

m(Ag₂CrO4) = 0.00860 mol Ag₂CrO4/331.728 g Ag₂CrO4/mol Ag₂CrO4 = 0.00003 g Ag₂CrO4 to one significant figure (due to 0.2 M and 0.3 M).