Question

We wish to determine how many gramsof potassium nitrate can form when 100.

mL of 0.40 M potassium chromate solution is added to excess silver nitrate.

2AgNO₃(aq) + K₂CrO₄ (aq) → Ag₂CrO₄(s) + 2KNO₃(aq)

How many moles of K₂CrO₄ are present in 100. mL of 0.40 M K₂CRO₄?

Answer 1

There are 0.040 moles of K₂CrO₄ in 100 mL of 0.40 M solution, resulting in the formation of 4.044 grams of KNO₃ when reacted with excess silver nitrate.

Step-by-step explanation:

To determine how many moles of K₂CrO₄ are present in 100 mL of 0.40 M K₂CrO₄, you can use the formula: moles = molarity × volume. Since the volume needs to be in liters for the calculation, convert 100 mL to 0.100 L (because 1000 mL = 1 L).

Then, multiply the molarity by the volume to find the moles of K₂CrO₄: moles = 0.40 mol/L × 0.100 L = 0.040 moles of K₂CrO₄. From the balanced chemical equation given, we know that 1 mole of K₂CrO₄ produces 2 moles of KNO₃. Therefore, the moles of KNO₃ formed will also be 0.040 moles.

To find the mass of KNO₃ created, use the molar mass of KNO₃ (101.1 g/mol). Mass = moles × molar mass, thus the mass of KNO₃ = 0.040 moles × 101.1 g/mol = 4.044 grams of KNO₃ formed.