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n=(1atm)(49L)/(0.0821)(273.15)=2.19 moles

2.19 moles x 44g of CO2 =96.36g of CO2




2.19 moles x 100g of CaCO3= 219g of CaCO3

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2009-11-06 07:51:52
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Q: How many grams of calcium carbonate are needed to produce 49.0 L of carbon dioxide at STP?
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