Could somebody tell me how i would go about solving this chemistry problem and what answer I should get?

Question

A seashell that is composed largely of calcium carbonate reacts with a solution of HCl. As a

result, 1500 mL of dry CO 2 gas at STP is produced. The other products are CaCl 2 and H 2 O.

a.) How many grams of CaCO 3 are consumed in the reaction?

b.) What volume of 2.00 M HCl solution is used in this reaction?

Answer 1

First, let's write the balanced equation CaCO3 + 2HCl ----> CaCl2 + H2O + CO2. Next let's calculate how many moles of CO2 are released. This is done by using the ideal gas law n = PV/RT. The pressure is one atmosphere, the volume 1.5 L, T = 273 C, R = 0.0821 L-atm/mol-deg. Solving for n gives 0.0669 moles CO2.

a. The conversion factor for CO2 to CaCO3 is 1 CaCO3/ 1 CO2 = 1/1 x 0.0669 moles CaCO3 for each mole of CO2. Therefore, 0.0669 moles x 100.086 g/mole = 6.70 g CaCO3.

b. The balanced equation indicates that 2 moles of HCl are needed for every mole of CO2 produced. So, 0.669 x 2 = 0.134 moles HCl needed. With a 2.00 molar HCl solution, you will need 1L/2.00 mole x 0.134 moles HCl = 0.067 L or 67 ml of 2.00 molar HCl.

Please check math.