Question

How many liters of carbon dioxide will be produced at STP if 3.56 g calcium carbonate reacts completely with carbon dioxide? CaCO3 --> CaO + CO2

Answer 1

Answer:

V = 0.798 L

Explanation:

Hello there!

In this case, for this gas stoichiometry problem, we first need to compute the moles of carbon dioxide via stoichiometry and the molar mass of starting calcium carbonate:

$3.56gCaCO_3*\frac{1molCaCO_3}{100gCaCO_3} *\frac{1molCO_2}{1molCaCO_3} =0.0356molCO_2$

Next, we use the ideal gas equation for computing the volume, by bearing to mind that the STP conditions stand for a pressure of 1 atm and a temperature of 273.15 K:

$PV=nRT\\\\V=\frac{nRT}{P}\\\\V=\frac{0.0356mol*0.08206\frac{atm*L}{mol*K}*273.15K}{1atm} \\\\V=0.798L$

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