mix 35.50 mL of a 0.201 M solution of K2CrO4 with 35.50 mL of a 0.201 M solution of AgNO3, what mass of solid will form?
K2CrO4 + 2AgNO3 ==> Ag2CrO4 + 2KNO3
moles K2CrO4 = M x L = ?
moles AgNO3 = M x L = ?
Now do two stoichiometry problems.
1. How many moles Ag2CrO4 if we used all of the K2CrO4. moles Ag2CrO4 = moles K2CrO4 x (1 mole Ag2CrO4/1 mole K2CrO4) = moles K2CrO4 x (1/1) = ?
2. How many moles Ag2CrO4 if we used all of the AgNO3? moles Ag2CrO4 = moles AgNO3 x (1 mole Ag2CrO4/2 moles AgNO3) = moles AgNO3 x (1/2) = ?
Calculate 1 gives you a different value than calculation 2. Both answers can't be right; the correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
Use the smaller value, convert to grams. g = moles x molar mass.
To find the mass of the solid that will form, we need to determine the limiting reagent first. The limiting reagent is the reactant that will be completely consumed and determine the maximum amount of product that can be formed.
Let's start by calculating the number of moles for each reactant:
Number of moles of K2CrO4 = volume (in liters) x molarity
= 0.03550 L x 0.201 mol/L
= 0.007 infinum.
Number of moles of AgNO3 = volume (in liters) x molarity
= 0.03550 L x 0.201 mol/L
= 0.007 infinum.
Now, let's compare the mole ratios of the reactants:
K2CrO4 : AgNO3 (from the balanced chemical equation)
1 : 2
Since the mole ratio of K2CrO4 to AgNO3 is 1:2, we can see that we have twice as many moles of AgNO3 as K2CrO4. Therefore, K2CrO4 is the limiting reagent.
To calculate the mass of the solid formed, we need to use the stoichiometry of the balanced chemical equation. The balanced equation for the reaction between K2CrO4 and AgNO3 is:
2 K2CrO4 + AgNO3 -> Ag2CrO4 + 2 KNO3
From the balanced equation, we can see that 2 moles of K2CrO4 react to form 1 mole of Ag2CrO4. Therefore, the mole ratio of K2CrO4 to Ag2CrO4 is 2:1.
The molar mass of Ag2CrO4 is calculated by adding up the atomic masses of the elements in the formula: Ag (2 x 107.87 g/mol) + Cr (1 x 52.00 g/mol) + O (4 x 16.00 g/mol) = 331.87 g/mol.
Now, we can calculate the mass of the solid formed:
Mass of Ag2CrO4 = number of moles of K2CrO4 x molar mass of Ag2CrO4
= 0.007 infinum x 331.87 g/mol
= 2.32 mg infinum.