(a) Calculate the mass of CaCO3 required to react completely with 25mL of 0.75 M HCl. (b) Calculate volume of CO2 released at STP in this reaction

Question

Answer nicely with the calculations showing by the side. In big or clear handwriting so it can be understandable. Thank u:)

Answer 1

Look at the balanced equation for the reaction takin place:

CaCO₃ + 2HCl ==> CaCl₂ + H₂O + CO₂

(a) moles HCl present = 25 ml x 1 L/1000 ml x 0.75 mol/L = 0.01875 mols HCl

mols CaCO₃ needed = 0.01875 mols HCl x 1 molCaCO₃ / 2 mol HCl = 0.009375 mols CaCO₃

mass CaCO₃ needed = 0.009375 mols CaCO₃ x 100.1 g CaCO₃/ mol = 0.938 g = 0.94 g CaCO₃ (2 sig. figs.)

(b) At STP 1 mole of any ideal gas = 22.4 L. Assuming ideal gas behavior..

mols CO₂ formed = 0.01875 mols HCl x 1 mol CO₂ / 2 mol HCl = 0.009375 mols CO₂

0.009375 mols CO₂ x 22.4 L / mol = 0.21 L CO₂