a) how many mL or carbon dioxide gas were generated by the decompostion of 6.24g of...
a) how many mL or carbon dioxide gas were generated by the decompostion of 6.24g of calcium carbonate at STP b) If 52.6 L of carbon dioxide at STP were needed, how many moles of calcium carbonate would be required?
Homework Answers
CaCO3 --> CaO + CO2
PV = nRT
mol of CaCO3 = mass/MW = 6.24/100 = 0.0624 mol
1 mol = 22.4L alwas at STP
0.0624 mol --> 0.0624 *22.4 = 1.39776 L
b.
mol of CO2 = 52.6/22.4 = 2.3482
ratio is 1:1
so
2.3482 mol of CaCO3 needed
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