Question

A reaction occurs between sodium bicarbonate and hydrochloric acid according to the equation NaHCO3(s) + HCl(aq) → CO₂(g) + H₂O(l) + NaCl(aq) If 24.50 mL of CO₂ is generated at a pressure of 660 torr and a temperature of 23.0°C, what mass of NaHCO3 reacted?​

Answer 1

To determine the mass of NaHCO3 reacted in the reaction between sodium bicarbonate (NaHCO3) and hydrochloric acid (HCl), we can use the ideal gas law to relate the volume of CO2 generated to the number of moles of CO2. Then, by considering the stoichiometry of the balanced chemical equation, we can calculate the corresponding moles of NaHCO3.

First, let's convert the given conditions to SI units:

Volume of CO2 = 24.50 mL = 0.02450 L

Pressure = 660 torr = 660/760 atm (since 1 atm = 760 torr)

Temperature = 23.0°C = 23.0 + 273.15 K

Next, we can use the ideal gas law equation:

PV = nRT

Where:

P = Pressure in atm

V = Volume in liters

n = Number of moles

R = Ideal gas constant (0.0821 L·atm/(mol·K))

T = Temperature in Kelvin

Rearranging the equation, we have:

n = PV / RT

Now, let's calculate the number of moles of CO2:

P = 660/760 atm

V = 0.02450 L

R = 0.0821 L·atm/(mol·K)

T = 23.0 + 273.15 K

n = (660/760) * (0.02450) / (0.0821) * (23.0 + 273.15)

n ≈ 0.001117 mol (approximately)

From the balanced chemical equation, we know that the stoichiometric ratio between NaHCO3 and CO2 is 1:1. Therefore, the number of moles of NaHCO3 reacted is also approximately 0.001117 mol.

The molar mass of NaHCO3 is:

Na = 22.99 g/mol

H = 1.01 g/mol

C = 12.01 g/mol

O = 16.00 g/mol (x3)

Total molar mass = 22.99 + 1.01 + 12.01 + (16.00 * 3) = 84.01 g/mol

Finally, we can calculate the mass of NaHCO3:

Mass = n * molar mass

Mass = 0.001117 mol * 84.01 g/mol

Mass ≈ 0.0938 g

Therefore, approximately 0.0938 grams of NaHCO3 reacted in the given reaction