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MariettaO [177]
4 years ago
14

When limestone rock, which is principally calcium carbonate, is heated, a reaction occurs. If 11.7 g of carbon dioxide were prod

uced in the lab from the decomposition of 30.7g of calcium carbonate, what is the percent yield for the reaction?
Chemistry
1 answer:
aev [14]4 years ago
6 0

Answer: 86.7 %

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

a) moles of CaCO_3

\text{Number of moles}=\frac{30.7g}{100g/mol}=0.307moles

CaCO_3\rightarrow CaO+CO_2

According to stoichiometry :

1 moles of CaCO_3 give = 1 mole of CO_2

Thus 0.307 moles CaCO_3 give =\frac{1}{1}\times 0.307=0.307moles of CO_2

Mass of CO_2=moles\times {\text {Molar mass}}=0.307moles\times 44g/mol=13.5g

\%\text{ yield }=\frac{\text{Actual yield}}{\text{Theoretical yield }}\times 100=\frac{11.7g}{13.5g}\times 100=86.7\%

Therefore, the percent yield for the reaction is, 86.7%

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Answer:

Percentage abundance of B 10 is = 20 %

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10.0129x+11.0093\left(100-x\right)=1081

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Answer: 0.0450 moles of CaCl_2

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According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number 6.023\times 10^{23} of particles.

To calculate the moles, we use the equation:

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