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A sample of 0.53 g of carbon dioxide was obtained by heating 1.31 $\mathrm{g}$ of calcalcium carbonate. What is the percennyield for this reaction?$\mathrm{CaCO}_{3}(s) \rightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(s)$
92.0$\%$
01:07
Sisi G.
Chemistry 101
Chapter 4
Stoichiometry of Chemical Reactions
Chemical reactions and Stoichiometry
University of Maryland - University College
Brown University
University of Toronto
Lectures
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for this next question, we need the balanced chemical reaction that is provided. We see that one mole of calcium carbonate produces one mole of calcium oxide and carbon dioxide. To determine the percent yield. We need to know what the theoretical yield of carbon dioxide will be. When 1.31 g of calcium carbonate decompose. We'll take the 1.31 g calcium carbonate converted into molds, calcium carbonate. By dividing by the molar mass calcium carbonate, 100.87 g. Then convert the moles calcium carbonate into molds. Carbon dioxide recognizing the strike geometry is a 1 to 1 mole relationship. Then, when we are in units of molds, carbon dioxide, we can calculate the theoretical amount, or the theoretical yield of carbon dioxide in grams by multiplying by the molar mass carbon dioxide And we get a theoretical yield of .5760 g. Carbon dioxide to calculate theoretical, I'm sorry to calculate percent yield. It's the theoretically, sorry, it's the actual yield Which is provided at 53g, divided by the theoretical yield that we just determined .5760 Times 100, so we get 92 on our percent yield.
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