Question

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Answer 1

Answer:

Option A is correct.

Pb(ClO3)2 was the limiting reagent, and Na2CrO4 remained in the solution.

Explanation:

The reaction is:

Pb(ClO₃)₂(aq) + Na₂CrO₄(aq) → PbCrO₄(s) + 2 NaClO₃(aq)

We determine the limiting reactant, so we need to convert the mass of the reactants, to moles:

88 g / 374.1 g/mol = 0.235 moles of Pb(ClO₃)₂

40.9 g / 162 g/mol = 0.252 moles of Na₂CrO₄

Ratio is 1:1 so, for 0.235 moles of Pb(ClO₃)₂, I need 0.235 moles of Na₂CrO₄ and for 0.252moles of Na₂CrO₄ I need 0.252 moles of Pb(ClO₃)₂.

We do not have enough Pb(ClO₃)₂ for the 0.252 moles of the sodium chromate, so the Pb(ClO₃)₂ is the limiting reactant and the Na₂CrO₄ is the excess reagent.

Answer 2

Answer:

Option A is correct.

Pb(ClO3)2 was the limiting reagent, and Na2CrO4 remained in the solution.

Explanation:

Step 1: Data given

Solution A = 88.0 grams Pb(ClO3)2

Molar mass Pb(ClO3)2 = 374.1 g/mol

Solution B = 40.9 grams of  Na2CrO4

 Molar mass Na2CrO4 = 161.97 g/mol

Step 2: The balanced equation

Pb(ClO3)2(aq) + Na2CrO4(aq) → PbCrO4(s) + 2 NaClO3(aq)

Step 3: Calculate moles

Moles = mass / molar mass

Moles Pb(ClO3)2 = 88.0 grams / 374.1 g/mol

Moles Pb(ClO3)2 = 0.235 moles

Moles Na2CrO4 = 40.9 grams / 161.97 g/mol

Moles Na2CrO4 = 0.253 moles

Step 4: Calculate the limiting reactant

For 1 mol Pb(ClO3)2 we need 1 mol Na2CrO4 to produce 1 mol PbCrO4 and 2 moles NaClO3

Pb(ClO3)2 is the limiting reactant. It will completely be consumed (0.235 moles). Na2CrO4 is in excess. There will react 0.235 moles. There will remain 0.253 - 0.235 = 0.018 moles

Option A is correct.

Pb(ClO3)2 was the limiting reagent, and Na2CrO4 remained in the solution.