Question

When aqueous solutions of K₂SO₄ and Pb(NO₃)₂ are combined, PbSO₄ precipitates. Calculate the mass, in grams, of the PbSO₄ produced when 1.66 mL of 0.2 M Pb(NO₃)₂ and 3.88 mL of 0.61 M K₂SO₄ are mixed. Calculate the mass to 3 significant figures.

Answer 1

Answer:

0.101g PbSO₄

Explanation:

To know an amount of a product we have to know first how many moles of reactants we have.

$1.66mLx0.2\frac{molPb(NO_{3})_{2}}{L}x\frac{1L}{1000mL}= 3.32x10^{-4}mol Pb(NO_{3})_{2}$

$3.88mLx\frac{0.61molK_{2}SO_{4}}{L}x\frac{1L}{1000mL} = 2.367x10^{-3}mol$$K_{2}SO_{4}$

The reaction is: $K_{2}SO_{4} + Pb(NO_{3})_{2}$⇒$PbSO_{4} + 2KNO_{3}$

We can observe that per one mole of K₂SO₄ we need one mole of Pb(NO₃)₂, so Pb(NO₃)₂ is the limitant reactant and reaction will stop when it´s over.

$3.32x10^{-4}mol Pb(NO_{3})_{2}x\frac{1molPbSO_{4}}{1molPb(NO_{3})_{2}} x\frac{303.26gPbSO_{4}}{1molPbSO_{4}}=0.101gPbSO_{4}$