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Multiple Choice
How many grams of silver chromate (Ag2CrO4) will precipitate when 150 mL of 0.5 M silver nitrate (AgNO3) is added to 100 mL of 0.4 M potassium chromate (K2CrO4)?
A
3.47 grams
B
6.94 grams
C
1.74 grams
D
13.88 grams
Verified step by step guidance
1
Write the balanced chemical equation for the reaction: 2 AgNO3 + K2CrO4 → Ag2CrO4 + 2 KNO3.
Calculate the moles of AgNO3 using the formula: moles = concentration (M) × volume (L). Convert 150 mL to liters and multiply by 0.5 M.
Calculate the moles of K2CrO4 using the formula: moles = concentration (M) × volume (L). Convert 100 mL to liters and multiply by 0.4 M.
Determine the limiting reactant by comparing the mole ratio from the balanced equation. For every 2 moles of AgNO3, 1 mole of Ag2CrO4 is formed. Compare the available moles of AgNO3 and K2CrO4 to find the limiting reactant.
Calculate the mass of Ag2CrO4 that will precipitate using the moles of the limiting reactant and the molar mass of Ag2CrO4. Use the formula: mass = moles × molar mass.