Question

Suppose you have a piece of aluminum that you want to react completely by the following single-displacement reaction: $$ 2 \mathrm{Al}(s)+6 \mathrm{HNO}_{3}(a q) \longrightarrow 2 \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}(a q)+3 \mathrm{H}_{2}(\mathrm{~g}) $$ (a) What mole ratio would you use in the following equation to determine the number of moles of \(\mathrm{HNO}_{3}\) needed to react with a known amount of Al? \(\operatorname{mol} \mathrm{Al} x\) \(=\mathrm{mol} \mathrm{HNO}_{3}\) (b) If you add more than enough nitric acid so that all the aluminum reacts, what mole ratio would you use in the following equation to determine the moles of \(\mathrm{H}_{2}\) produced? $$ \mathrm{mol} \mathrm{Al} \times==\mathrm{mol} \mathrm{H}_{2} $$ (c) Suppose you know the number of moles of \(\mathrm{H}_{2}\) product formed and you want to know the number of moles of Al that reacted. What mole ratio would you use in the following equation? $$ \mathrm{mol} \mathrm{H}_{2} \times \overline{\mathrm{mol}} \mathrm{Al} $$

Short answer

(a) The mole ratio of \(\mathrm{HNO_{3}}\) to \(\mathrm{Al}\) is 3:1. \n(b) The mole ratio of \(\mathrm{H_{2}}\) to \(\mathrm{Al}\) is 1.5:1. \n(c) The mole ratio of \(\mathrm{Al}\) to \(\mathrm{H_{2}}\) is 0.67:1.

Step-by-step solution

Calculation of Mole Ratio for Nitric Acid and Aluminum

To find the mole ratio of nitric acid to aluminium, refer to the chemical equation: \(2 \mathrm{Al}(s)+6 \mathrm{HNO}_{3}(a q) \rightarrow 2 \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}(a q)+3 \mathrm{H}_{2}(\mathrm{~g}) \). The coefficient of \(\mathrm{Al}\) is 2 and the coefficient of \(\mathrm{HNO_{3}}\) is 6. The ratio of \(\mathrm{Al}\) to \(\mathrm{HNO_{3}}\) is therefore \(\frac{2}{6} = \frac{1}{3}\). So, for every 1 mole of \(\mathrm{Al}\), 3 moles of \(\mathrm{HNO_{3}}\) are needed.

Calculation of Mole Ratio for Hydrogen and Aluminum

Refer once again to the chemical equation for the reaction. The coefficient of \(\mathrm{Al}\) is 2 and the coefficient of \(\mathrm{H_{2}}\) is 3. The ratio of \(\mathrm{Al}\) to \(\mathrm{H_{2}}\) is therefore \(\frac{2}{3}\). So, for every 1 mole of \(\mathrm{Al}\), 1.5 moles of \(\mathrm{H_{2}}\) are produced.

Calculation of Mole Ratio for Aluminum and Hydrogen gas

Once again referring to the chemical equation for the reaction, the coefficient of \(\mathrm{Al}\) is 2 and the coefficient of \(\mathrm{H_{2}}\) is 3. The ratio of \(\mathrm{H_{2}}\) to \(\mathrm{Al}\) is therefore \(\frac{3}{2}\). Thus, for every 1 mole of \(\mathrm{H_{2}}\), \(\frac{2}{3}\) moles of \(\mathrm{Al}\) would have been reacted.

Key concepts

Mole Ratio

The mole ratio is a crucial aspect of stoichiometry, which is the study of the quantitative relationships between the reactants and products in a chemical reaction. This concept helps us understand the proportions in which chemicals react.

For instance, in the provided exercise, we're looking at a single-displacement reaction involving aluminum (Al) and nitric acid (HNO3), leading to the production of aluminum nitrate and hydrogen gas. To find the mole ratio for Al to HNO3, we look at the coefficients in the balanced chemical equation provided. For every 2 moles of Al, 6 moles of HNO3 are required, giving us a mole ratio of 1:3 for Al:HNO3. This is fundamental in determining how much HNO3 is needed for a known amount of Al.

Single-Displacement Reaction

In a single-displacement reaction, an element reacts with a compound and takes the place of another element within that compound.

Our exercise showcases a single-displacement reaction where aluminum displaces hydrogen from nitric acid. This type of reaction is common in metallurgy and industrial processes and is characterized by its straight-forward predictability, as elements from the activity series displace less active ones. Understanding these reactions allows us to predict the products and calculate the amounts of reactants needed or products formed.

Chemical Equation

A chemical equation is a symbolic representation of a chemical reaction, displaying the reactants and the products with their respective chemical formulas. The numbers in front of the chemical formulas are coefficients and they depict the necessary proportions of each substance.

For example, the equation given in the exercise, \[2 \mathrm{Al}(s) + 6 \mathrm{HNO}_{3}(aq) \longrightarrow 2 \mathrm{Al}(\mathrm{NO}_{3})_{3}(aq) + 3 \mathrm{H}_{2}(g)\], consists of reactants (aluminum and nitric acid) and products (aluminum nitrate and hydrogen gas). The coefficients (2 for Al, 6 for HNO3, etc.) are the keys to understanding the mole ratios.

Reaction Stoichiometry

Reaction stoichiometry deals with the quantitative relationship between reactants and products in a chemical reaction. It's essential for predicting the yields of reactions and for scaling them up or down for practical use.

In the context of our exercise, by knowing the mole ratios from the balanced equation, we can calculate how much of each reactant will be consumed and how much of each product will be formed. For instance, with a known amount of Al, we can use the mole ratio (1 mole of Al yields 1.5 moles of H2) to find the amount of hydrogen gas produced when Al reacts with an excess of nitric acid. Understanding these ratios is fundamental for everything from laboratory experiments to industrial chemical production.