Question

HELP!!!! Will name brainliest
A precipitate of lead(II) chromate can be formed using the reaction below.

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

A chemist prepared the following two solutions and mixed them together in a large flask.
Solution A: 88.0 grams of solid Pb(ClO3)2 dissolved in 200 mL of water
Solution B: 40.9 grams of solid Na2CrO4 dissolved in 350 mL of water

What was the limiting reagent in the reaction that occurred when the solutions were mixed together, and which reagent remained in the solution after the reaction had occurred?
A
Pb(ClO3)2 was the limiting reagent, and Na2CrO4 remained in the solution.


B
Na2CrO4 was the limiting reagent, and Na2CrO4 remained in the solution.


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


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

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.