Question

Nitrogen gas reacts with hydrogen gas to produce ammonia by the following equation: $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) $$ a. If you have \(3.64 \mathrm{~g}\) of \(\mathrm{H}_{2}\), how many grams of \(\mathrm{NH}_{3}\) can be produced? b. How many grams of \(\mathrm{H}_{2}\) are needed to react with \(2.80 \mathrm{~g}\) of \(\mathrm{N}_{2}\) ? c. How many grams of \(\mathrm{NH}_{3}\) can be produced from \(12.0 \mathrm{~g}\) of \(\mathrm{H}_{2}\) ?

Short answer

a. 20.49 g of \text{NH}_3b. 0.6048 g of \text{H}_2c. 67.58 g of \text{NH}_3

Step-by-step solution

Calculate the molar mass of each compound

The molar mass of nitrogen gas, \text{N}_2, is calculated as follows: \( \text{M}(\text{N}_2)=2 \times 14.01 \text{g/mol} = 28.02 \text{g/mol} \).The molar mass of hydrogen gas, \text{H}_2, is: \( \text{M}(\text{H}_2)=2 \times 1.008 \text{g/mol} = 2.016 \text{g/mol} \).The molar mass of ammonia, \text{NH}_3, is: \( \text{M}(\text{NH}_3)=14.01 + 3 \times 1.008 \text{g/mol} = 17.034 \text{g/mol} \).

Solve Part (a)

Given mass of \text{H}_2 = 3.64 \text{g}.Convert mass of \text{H}_2 to moles: \[ \text{moles of H}_2 = \frac{3.64 \text{g}}{2.016 \text{g/mol}} = 1.805 \text{mol} \].According to the balanced equation, 3 moles of \text{H}_2 produce 2 moles of \text{NH}_3.So, 1.805 moles of \text{H}_2 will produce: \[ \text{moles of NH}_3 = 1.805 \text{mol H}_2 \times 2 \text{mol NH}_3 / 3 \text{mol H}_2 = 1.203 \text{mol NH}_3 \].Convert moles of \text{NH}_3 to grams: \[ \text{mass of NH}_3 = 1.203 \text{mol} \times 17.034 \text{g/mol} = 20.49 \text{g} \].

Solve Part (b)

Given mass of \text{N}_2 = 2.80 \text{g}.Convert mass of \text{N}_2 to moles: \[ \text{moles of N}_2 = \frac{2.80 \text{g}}{28.02 \text{g/mol}} = 0.1 \text{mol} \].According to the balanced equation, 1 mole of \text{N}_2 reacts with 3 moles of \text{H}_2.So, 0.1 moles of \text{N}_2 will require: \[ \text{moles of H}_2 = 0.1 \text{mol N}_2 \times 3 \text{mol H}_2 / 1 \text{mol N}_2 = 0.3 \text{mol H}_2 \].Convert moles of \text{H}_2 to grams: \[ \text{mass of H}_2 = 0.3 \text{mol} \times 2.016 \text{g/mol} = 0.6048 \text{g} \].

Solve Part (c)

Given mass of \text{H}_2 = 12.0 \text{g}.Convert mass of \text{H}_2 to moles: \[ \text{moles of H}_2 = \frac{12.0 \text{g}}{2.016 \text{g/mol}} = 5.952 \text{mol} \].According to the balanced equation, 3 moles of \text{H}_2 produce 2 moles of \text{NH}_3.So, 5.952 moles of \text{H}_2 will produce: \[ \text{moles of NH}_3 = 5.952 \text{mol H}_2 \times 2 \text{mol NH}_3 / 3 \text{mol H}_2 = 3.968 \text{mol NH}_3 \].Convert moles of \text{NH}_3 to grams: \[ \text{mass of NH}_3 = 3.968 \text{mol} \times 17.034 \text{g/mol} = 67.58 \text{g} \].

Key concepts

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. For example, nitrogen gas reacts with hydrogen gas to form ammonia. This reaction is represented by the equation: ewline $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) $$ ewline The substances on the left side, nitrogen and hydrogen, are the reactants. Ammonia on the right side is the product. Understanding chemical reactions is crucial because they explain how substances change and interact. ewline In chemical reactions, bonds between atoms in the reactants are broken and new bonds are formed in the products. ewline During this process, the number of atoms remains the same, adhering to the Law of Conservation of Mass.

Mole Calculations

Mole calculations allow chemists to count particles by weighing them. A mole represents Avogadro's number, about 6.022 x 10^23 particles. For the given exercise, to find out how many grams of a substance are required or produced, you need to convert between grams and moles. ewline For example, if you have 3.64 grams of hydrogen (\text{H}_2), convert this to moles using its molar mass: ewline $$ \text{Moles of H}_2 = \frac{3.64 \text{g}}{2.016 \text{g/mol}} = 1.805 \text{mol} $$ ewline Moles help determine the relationships between different substances in a reaction, as indicated by the balanced chemical equation.

Molar Mass

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). Each element has a unique molar mass, found on the periodic table. ewline To calculate the molar mass of molecules, sum the molar masses of the constituent atoms. For example: ewline $$ \text{M}(\text{N}_2) = 2 \times 14.01 = 28.02 \text{g/mol} $$ ewline $$ \text{M}(\text{H}_2) = 2 \times 1.008 = 2.016 \text{g/mol} $$ ewline $$ \text{M}(\text{NH}_3) = 14.01 + 3 \times 1.008 = 17.034 \text{g/mol} $$ ewline These values are crucial for converting between grams and moles in stoichiometric calculations.

Balanced Equations

A balanced chemical equation has equal numbers of each type of atom on both sides. This ensures the Law of Conservation of Mass is followed. For instance, in the ammonia formation reaction: ewline $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) $$ ewline The equation shows that 1 mole of nitrogen gas (\text{N}_2) reacts with 3 moles of hydrogen gas (\text{H}_2) to produce 2 moles of ammonia (\text{NH}_3). ewline Balancing equations involves adjusting coefficients, not subscripts, to ensure the same number of each atom on both sides. This step is critical for accurate stoichiometric calculations, as it shows the exact proportions in which reactants combine to form products.

Mass-to-Mass Conversion

Mass-to-mass conversion in stoichiometry involves converting the mass of a reactant to the mass of a product. This requires several steps using molar masses and the balanced chemical equation. ewline For example, to convert how much ammonia (\text{NH}_3) we can produce from 3.64 grams of hydrogen (\text{H}_2): ewline $$ \text{Step 1: Convert mass of H}_2 \text{ to moles:} $$ ewline $$ \text{moles of H}_2 = \frac{3.64 \text{g}}{2.016 \text{g/mol}} = 1.805 \text{mol} $$ ewline $$ \text{Step 2: Use the balanced equation to find moles of NH}_3\text{ produced:} $$ ewline $$ \text{moles of NH}_3 = 1.805 \text{mol H}_2 \times \frac{2 \text{mol NH}_3}{3 \text{mol H}_2} = 1.203 \text{mol NH}_3 $$ ewline $$ \text{Step 3: Convert moles of NH}_3 \text{ to grams:} $$ ewline $$ \text{mass of NH}_3 = 1.203 \text{mol} \times 17.034 \text{g/mol} = 20.49 \text{g} $$ ewline This technique is essential for predicting the amounts of products formed in reactions based on given quantities of reactants.