Question

A 151.10 mL volume of hydrochloric acid reacts completely with a solid sample of MgCO3. The reaction is given below.

2 HCl(aq) + MgCO3(s) → CO2(g) + H2O(l) + MgCl2(aq)

The volume of CO2 formed is 153.00 mL at 27.1 °C and 772 mmHg. What is the molarity of the HCl solution?

Answer 1

0.0415 M

Step-by-step explanation:

To solve this problem, we need to use the ideal gas law to calculate the number of moles of CO2 formed, and then use stoichiometry to calculate the number of moles of HCl used in the reaction. Finally, we can calculate the molarity of the HCl solution using the volume and number of moles of HCl.

First, let's calculate the number of moles of CO2 formed:

PV = nRT

where P is the pressure in atm, V is the volume in L, n is the number of moles, R is the gas constant (0.08206 L·atm/K·mol), and T is the temperature in Kelvin.

We need to convert the volume of CO2 from mL to L, the temperature from Celsius to Kelvin, and the pressure from mmHg to atm:

V = 153.00 mL = 0.15300 L

T = 27.1 °C + 273.15 = 300.25 K

P = 772 mmHg / 760 mmHg/atm = 1.016 atm

Plugging in these values, we get:

n = PV/RT = (1.016 atm)(0.15300 L)/(0.08206 L·atm/K·mol)(300.25 K) ≈ 0.00627 mol CO2

According to the balanced chemical equation, the reaction between HCl and MgCO3 has a 1:1 stoichiometry, which means that the number of moles of HCl used in the reaction is also approximately 0.00627 mol.

The volume of HCl used is 151.10 mL, or 0.15110 L. Therefore, the molarity of the HCl solution is:

M = n/V = 0.00627 mol / 0.15110 L ≈ 0.0415 M

Therefore, the molarity of the HCl solution is approximately 0.0415 M.