Question

For the following chemical equation:

CaCO3(s) + 2HCl(aq).......>H2)(l) +CO2 (g) +CaCl2 (aq)

A). How many liters of CO2 can form at STP when 18.5 mL of a 3.51 M HCl solution reacts with excess CaCO3?

B). What is the molarity of a HCl solution if the reaction of 210 mL of the HCl solution with excess CaCO3 produces 12.1 L of CO2 gas at 725 mmHg and 18 C?

Answer 1

A). How many liters of CO₂ can form at STP when 18.5 mL of a 3.51 M HCl solution reacts with excess CaCO₃?

CaCO₃(s) + 2HCl(aq)H2(l) +CO₂ (g) +CaCl₂ (aq)

Since CaCO₃ is in excess we need to calculate number of moles of HCl.

18.5 mL of 3.51 M means = 3.51 x 18.5/1000 = 0.0649 moles

Two moles of HCl will generate 1 mole of CO₂ as per the reaction

so 0.0649 moles will generate 0.0324 moles of CO₂

Standard temperature and pressure (informally abbreviated as STP) as a temperature of 273.15 K (0 °C, 32 °F) and an absolute pressure of exactly 100,000 Pa (1 bar, 14.5 psi, 0.98692 atm).

so using PV = nRT

we calculate V = nRT/P = 0.0324 mol x 8.314 L kPa K⁻¹ mol⁻¹x 273.15 K/100 kPa

V = 0.737 L of CO₂

B) What is the molarity of a HCl solution if the reaction of 210 mL of the HCl solution with excess CaCO₃ produces 12.1 L of CO2 gas at 725 mmHg and 18 C?

First we calculate moles of CO₂

725 millimeter of mercury = 96.66 kPa

18 C = 291 K

so using PV = nRT

n = PV/RT = 96.66kPa x 12.1L/8.314 L kPa K⁻¹ mol⁻¹ x 291K

=(1169/2419) mol

=0.483 mol of CO₂ is produced

As per the equation the amount of HCl required should be double the CO₂ produced

So HCl required would be 0.967 moles

Since we have used 210 mL the molarity of the solution should be 4.60 M