0.196 M lead (II) nitrate was reacted with 0.277 M potassium carbonate as follows: Pb(NO3)2(aq) +...

Transcribed Image Text:

0.196 M lead (II) nitrate was reacted with 0.277 M potassium carbonate as follows: Pb(NO3)2(aq) + K2CO3(aq) PbCO3(s) + 2 KNO3(aq) Determine the percent yield if 242.5 mL of each reactant were allowed to react, and a mass of 7.757 g of solid were obtained. Note: Do not use scientific notation or units in your response. Sig figs will not be graded in this question, enter your response to four decimal places. Carmen may add or remove digits from your response, your submission will still be graded correctly if this happens. 0.196 M lead (II) nitrate was reacted with 0.277 M potassium carbonate as follows: Pb(NO3)2(aq) + K2CO3(aq) PbCO3(s) + 2 KNO3(aq) Determine the percent yield if 242.5 mL of each reactant were allowed to react, and a mass of 7.757 g of solid were obtained. Note: Do not use scientific notation or units in your response. Sig figs will not be graded in this question, enter your response to four decimal places. Carmen may add or remove digits from your response, your submission will still be graded correctly if this happens.

Answer rating: 100% (QA)

To determine the percent yield of the reaction we first need to calculate the theoretical yield of l... View the full answer