Question

A solution of potassium chromate reacts with a solution of lead(II) nitrate to produce a yellow precipitate of lead(II) chromate and a solution of potassium nitrate.
(a) Write the balanced chemical equation. (Use the lowest possible coefficients. Include states-of-matter under the given conditions in your answer.)
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(b) Starting with 0.130 mol potassium chromate, determine the mass of lead chromate that can be formed.
g

Answer 1

Answer:

a) K₂CrO₄ (aq) + Pb(NO₃)₂ (aq) → 2KNO₃ (aq) + PbCrO₄ (s)

b) $m (PbCrO_{4}) = 42.0g$

Explanation:

a) This reaction can be classified as a precipitation reaction as the product was precipitated due to strong ionic bond between lead (II) and chromate ions.

b) This part can be solved by stoichiometric calculations.

From balanced chemical equation it can be seen that 1 mol of potassium chromate gives 1 mol of lead chromate (i.e. they have 1:1 mole ratio), which equals to 323.194 g. So by applying the ratio we can say that 0.130 mol of potassium chromate will give:

$m (PbCrO_{4}) = \frac{(323.194 g)(0.130 mol)}{1 mol}$

$m (PbCrO_{4}) = 42.0 g$