Calculate CaCO3 Mass: Stoichiometry Help for Decomposition Reaction

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A sample of CaCO3 decomposes upon heating, releasing 1.5432 g of CO2, which allows for the calculation of the original mass of CaCO3. The calculation shows that 3.5097 g of CaCO3 was present in the original sample. To find the percentage by mass of CaCO3 in a 5.768 g sample, the correct method is to divide 3.5097 g by 5.768 g and multiply by 100. This approach confirms the accuracy of the calculations presented. The discussion emphasizes the stoichiometric relationships in the decomposition reaction.
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a. A sample of CaCO3 decomposes when heated to form calcium oxide and carbon dioxide. 1.5432 g of CO2 are released in the reaction. Find g of CaCO3 in the original sample.
CaCO3 --> CO2 + CaO

1.5432 g CO2*(1 mol of CO2/44.01 g) = 0.035065 mol of CO2*(1 mol CaCO3/1 mol CO2) = 0.035065 mol CaCO3

0.035065 mol CaCO3*(100.09 g/ 1mol CaCO3) = 3.5097 g CaCO3??


b. The original sample was a mixture from which only CaCO3 released carbon dioxide. Calculate the percentage by mass of CaCO3 if original's sample mass was 5.768 g.


Do I divide 3.5097 g CaCO3/5.768 g * 100 or must I subtract the grams of CO2 from the original sample's mass?

Thanks.
 
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Any volunteers?

Thanks again.
 
Soaring Crane said:
a. A sample of CaCO3 decomposes when heated to form calcium oxide and carbon dioxide. 1.5432 g of CO2 are released in the reaction. Find g of CaCO3 in the original sample.
CaCO3 --> CO2 + CaO

1.5432 g CO2*(1 mol of CO2/44.01 g) = 0.035065 mol of CO2*(1 mol CaCO3/1 mol CO2) = 0.035065 mol CaCO3

0.035065 mol CaCO3*(100.09 g/ 1mol CaCO3) = 3.5097 g CaCO3??
That's correct.


b. The original sample was a mixture from which only CaCO3 released carbon dioxide. Calculate the percentage by mass of CaCO3 if original's sample mass was 5.768 g.


Do I divide 3.5097 g CaCO3/5.768 g * 100 or must I subtract the grams of CO2 from the original sample's mass?

Thanks.
This looks a little silly, but your first idea seems right.
 
all seemed right: (3.5097/5.768)*100 should do it.
 
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