Question

CaCO3(s)→CaO(s)+CO2(g)

What is the mass of calcium carbonate needed to produce 37.0 L of carbon dioxide at STP?

Answer 1

1 mole of calcium carbonate produces one mole of CO2 i.e 22.4L volume of CO2

moles of CaCO3 required to produce 37 litre of CO2=37/22.4

=370/224

=185/112

=1.7 moles(approx)

mass of one mole=40+12+48=100g

mass of 1.7 mole=1700g...(approx)

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Answer 2

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$</p><p>\underline {\underline {\tt{\oramge{Step-By-Step-Explaination \: : }}}}}</p><p>$

At STP, 1 mole of a gas weighs 22.4 L

Here, Anmount of Carbon Dioxide gas produced is 37.0 L.

Amount of gas required = 27/16 mole CO2

1 mole of CaCo3 gives 1 mole of CO2 gas.

27/16 mole of CaCo3 gives 27/16 mole of CO2

No of moles of CaCo3 required = 27/16

Given Mass = No of Moles × Molar Mass = 27/16 × 100 grams = 168.75 grams.