For the chemical reaction 2AgNO3+Na2CrO4⟶Ag2CrO4+2NaNO3 how many moles of silver chromate will be produced from 144 g of silver nitrate?

Answer 1

2AgNO₃ + Na₂CrO₄ ==> Ag₂CrO₄ + 2NaNO₃ ... balanced equation

molar mass AgNO₃ = 169.9 g / mol

moles AgNO₃ used = 144 g x 1 mol / 169.9 g = 0.8476 moles

Assuming Na₂CrO₄ is present in excess, we can use dimensional analysis and the stoichiometry of the balanced equation to find moles of product, Ag₂CrO₄ formed.

0.8476 mols AgNO₃ x 1 mol Ag₂CrO₄ / 2 mols AgNO₃ = 0.424 mols Ag₂CrO₄ (3 sig.figs.)

Answer 2

1. Calculate the molecular mass of AgNO3 in grams (call this x)
2. Find the moles of AgNO3 used = 144/x (call this y)
3. From the equation for every 2 moles of AgNO3 you get 1 mole of silver chromate so number of moles of silver chromate will be (y/2)