Question

1a) Balance this equation:

CaCO3+HCl—>CaCl2+H2O+CO2

1b) Calculate the mass of pure calcium carbonate needed to produce 40 mL of CO2 at 1 atm and 25 degrees Celsius

Answer 1

Here is the Solution:

119]. Balance this equation. CaCO3 + HCl - CaCl2 + H₂O + CO2 ► to balance the equation, first we will count the number of atoms in each element on the left hand side and right hand side of the reaction we get, Left hand side Right hand side There is 1 hydrogen (H) and 1 chlorine (61) on the left hand side but 2 on the right hand side so, this is unbalanced, so therefore we need to balance it by using 2 molecules of Hel. We get , CaCO3 + 2HCl CaCl2 + H2O + CO2 This is the balanced equation * 2b We need to calccelate the pure calcium carbonate (CaCO3) needed to produce 40 mL of co2 at i atm and 25°C first we will calculate the moles of CO2 gas in 40mL using ideal gas law : PVC DRT 40 mL is equals to 0.040 L n = PV = (1 atm) x(0.04045 1 (0.0821 Kath /mol KSX/298. Ich = 0.001634 mol CO₂ т? AT (0.0621 . 4m/s means, this will require 0.001634 mol CaCO3 we know that, the molar mass of CaCO3 is 100.0869 g/mol So, mass caCO3 = 0.001634 not x 100.0869 g/mot mass calO3 = 10. 1635 g needed.

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