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

V = 0.798 L

Step-by-step 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*(1molCaCO_3)/(100gCaCO_3) *(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=(nRT)/(P)\\\\V=(0.0356mol*0.08206(atm*L)/(mol*K)*273.15K)/(1atm) \\\\V=0.798L`

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