Question

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

The volume of carbon dioxide gas produced is 0.106 liters.

Step-by-step solution

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.

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.

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.

Key concepts

Understanding Stoichiometry

Stoichiometry is like a recipe for chemical reactions. It helps us understand the relationships between reactants and products. When you bake cookies, you follow a recipe that tells you exactly how much of each ingredient you need. Similarly, in a chemical reaction, stoichiometry tells us how much of each substance is involved.

In the exercise, we used stoichiometry to find out how much carbon dioxide ( CO_2 ) is produced from a specific amount of sodium bicarbonate ( NaHCO_3 ).

• The balanced chemical equation shows the ratio of the substances. Here, 1 mole of NaHCO_3 produces 1 mole of CO_2 .
• By knowing the moles of NaHCO_3 , we could directly find the moles of CO_2 .

This is the magic of stoichiometry. It allows us to use known quantities to discover unknowns in a reaction.

Exploring Chemical Reactions

Chemical reactions are processes where substances, known as reactants, are transformed into different substances, called products. They involve breaking and forming bonds between atoms.

In our example, the reaction between antacid ( NaHCO_3 ) and hydrochloric acid ( HCl ) produces salt ( NaCl ), water ( H_2O ), and carbon dioxide ( CO_2 ).

• Reactants: NaHCO_3 and HCl .
• Products: NaCl , H_2O , and CO_2 .

This balanced equation ensures that the number of atoms for each element is the same on both sides. It's like maintaining a balance, where nothing is lost or gained, just rearranged.

Molar Mass Calculation Explained

Molar mass is a bridge between grams and moles. It’s the weight of one mole of a substance, measured in grams per mole (\text{g/mol}). Calculating molar mass involves adding up the atomic masses from the periodic table.

For sodium bicarbonate (NaHCO_3):

• Sodium (Na) is 22.99 \text{g/mol}.
• Hydrogen (H) is 1.01 \text{g/mol}.
• Carbon (C) is 12.01 \text{g/mol}.
• Oxygen (O) is 16.00 \text{g/mol} (there are three oxygens, so multiply by 3).

The total molar mass is 84.01 \text{g/mol}. This allows us to convert a given mass of a substance (like 0.350 g of NaHCO_3) into moles, using: \[\text{Number of moles} = \frac{\text{Mass}}{\text{Molar Mass}}\]. Understanding molar mass is essential for predicting and measuring amounts in chemical reactions.