Answer to Question #167946 in Chemistry for Ava Brooke

Question

 After it had cooled the dry marble and test tube weighed 17.56 g. I know it is difficult to see the tick marks for the volume of gas in the video so by the end I collected 35.3 mL of CO2. Calculate the moles of calcium carbonate (marble) reacted (use 3 decimals)

Answer 1

CaCO₃ + 2HCl = H₂O + CO₂ + CaCl₂

n (CO₂) = n (CaCO₃)

n (CO₂) = n (CaCO₃) = V/22.4 = 0.0353/22.4 = 0.002 mol