Question

A chemistry student wants to perform a reaction that produces carbon dioxide gas. If the student has 500 grams of calcium carbonate and an excess of hydrochloric acid, how many liters of carbon dioxide gas can be produced at Standard Temperature and Pressure (STP)? (Assume the reaction goes to completion)

Answer 1

To solve this problem, we will use the balanced chemical equation for the reaction between calcium carbonate and hydrochloric acid, the molar mass of calcium carbonate, and the molar volume of a gas at STP.The balanced chemical equation for the reaction is:CaCO3  s  + 2 HCl  aq   CaCl2  aq  + H2O  l  + CO2  g First, we need to find the moles of calcium carbonate  CaCO3  in 500 grams. The molar mass of CaCO3 is:Ca = 40.08 g/molC = 12.01 g/molO = 16.00 g/molMolar mass of CaCO3 = 40.08 + 12.01 +  3  16.00  = 100.09 g/molNow, we can find the moles of CaCO3:moles of CaCO3 = mass / molar mass = 500 g / 100.09 g/mol = 4.995 molesFrom the balanced chemical equation, we can see that 1 mole of CaCO3 produces 1 mole of CO2. Therefore, 4.995 moles of CaCO3 will produce 4.995 moles of CO2.At STP  Standard Temperature and Pressure: 0C and 1 atm , 1 mole of any gas occupies 22.4 liters. So, we can find the volume of CO2 produced:Volume of CO2 = moles of CO2  molar volume at STP = 4.995 moles  22.4 L/mol = 111.89 LTherefore, the student can produce approximately 111.89 liters of carbon dioxide gas at STP.