Question

0.329 M copper (II) nitrate was reacted with 0.528 M potassium carbonate as follows: Cu (NO subscript 3 )subscript 2 italic (a q italic )space plus space straight K subscript 2 CO subscript 3 italic (a q italic )rightwards arrow CuCO subscript 3 italic (s italic )space plus space 2 thin space KNO subscript 3 italic (a q italic )Determine the percent yield if 163.9 mL of each reactant were allowed to react, and a mass of 4.883 g of solid were obtained.

Answer 1

Answer:

73.33% is the percent yield

Explanation:

Percent yield is defined as:

Actual yield (4.883g) / Theoretical yield * 100

Based on the reaction:

Cu(NO₃)₂(aq) + K₂CO₃(aq) → CuCO₃(s) + 2KNO₃(aq)

1 mole of copper nitrate reacts per mol of potassium carbonate.

To solve this question we must find limiting reactant. With limiting reactant we can find the theoretical moles of solid produced and its mass as follows:

Moles Cu(NO₃)₂:

0.1639L * (0.329mol / L) = 0.0539 moles

Moles K₂CO₃:

0.1639L * (0.528mol / L) = 0.0865 moles

As the reaction is 1:1, the limiting reactant is Cu(NO₃)₂.

1 mol of Cu(NO₃)₂ produces 1 mol of CuCO₃. That means theoretical moles produced are 0.0539 moles. And the mass is:

Mass CuCO₃ -Molar mass: 123.55g/mol-

0.0539 moles * (123.55g / mol) = 6.659g of CuCO₃ is the theoretical mass

And percent yield:

4.883g / 6.659g * 100

73.33% is the percent yield