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Chapter 16: Problem 84

A typical reaction between an antacid and the hydrochloric acid in gastric juice is \(\mathrm{NaHCO}_{3}(a q)+\mathrm{HCl}(a q) \longrightarrow\) $$\mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)$$ Calculate the volume (in liters) of \(\mathrm{CO}_{2}\) generated from \(0.350 \mathrm{~g}\) of \(\mathrm{NaHCO}_{3}\) and excess gastric juice at \(1.00 \mathrm{~atm}\) and \(37.0^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The volume of carbon dioxide gas produced is 0.106 liters.

Step by step solution

01

Calculate moles of Sodium Bicarbonate

First, calculate the molar mass of \(NaHCO_3\). Adding up the atomic masses of Na (22.99 g/mol), H (1.01 g/mol), C (12.01 g/mol) and O (16.00 g/mol), we get 84.01 g/mol. Since we know the mass is 0.350 g, to get the number of moles we divide the mass by the molar mass. Number of moles = \(\frac{0.350}{84.01} = 0.00417\) moles.
02

Use stoichiometry

From the balanced chemical reaction, we know that one mole of sodium bicarbonate (\(NaHCO_3\)) reacts to generate one mole of carbon dioxide (\(CO_2\)). Therefore, number of moles of \(CO_2\) produced = number of moles of \(NaHCO_3\) = 0.00417 moles.
03

Use the ideal gas law

The ideal gas law is \(PV = nRT\), where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. We are given P = 1.00 atm, n = 0.00417 moles, R = 0.0821 L.atm/mol.k (standard value), and T = 37.0°C which we need to convert to Kelvin by adding 273, T = 37.0 + 273 = 310 K. Solving for V, we have \(V = \frac{nRT}{P} = \frac{(0.00417 mol)(0.0821 L.atm/mol.k)(310 K)}{1.00 atm} = 0.106\) L.

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