Consider the unbalanced chemical equation \(\mathrm{CaCN}_{2}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{CaCO}_{3}+\mathrm{NH}_{3}\) If you began the reaction with \(5.65 \mathrm{~g}\) of \(\mathrm{CaCN}_{2}\) and \(12.2 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\), how many grams of \(\mathrm{NH}_{3}\) would be produced if the reaction had an \(86.0 \%\) yield?

To balance the equation, we need to make sure that the number of atoms of each element on the reactants side is equal to the number of atoms of each element on the products side. The balanced chemical equation: \( CaCN_2 + 2H_2O \rightarrow CaCO_3 + 2NH_3 \) Now, the equation is balanced, and we can proceed to finding the limiting reactant.

To find the limiting reactant, we will first find the molar mass of each reactant and convert grams to moles for each. Next, we will use the mole ratio (stoichiometry) from the balanced chemical equation to determine the limiting reactant. Molar masses: - CaCN2: Ca (40.08 g/mol) + 2C (24.02 g/mol) + 2N (28.02 g/mol) = 92.12 g/mol - H2O: 2H (2.02 g/mol) + O (16.00 g/mol) = 18.02 g/mol Moles of reactants: - Moles of CaCN2: 5.65 g / 92.12 g/mol ≈ 0.0613 mol - Moles of H2O: 12.2 g / 18.02 g/mol ≈ 0.677 mol Now, we will use the mole ratio in the balanced equation to determine the limiting reactant. As the ratio is 1 mole of CaCN2 : 2 moles of H2O, we will divide the number of moles of each reactant by the stoichiometric coefficients. - CaCN2: 0.0613 mol / 1 = 0.0613 - H2O: 0.677 mol / 2 ≈ 0.3385 Since 0.0613 is smaller than 0.3385, CaCN2 is the limiting reactant.

Now that we know CaCN2 is the limiting reactant, we can calculate the theoretical yield of NH3 using the mole ratio. The ratio in the balanced equation is 1 mole of CaCN2 : 2 moles of NH3. To find how many moles of NH3 would be produced, we have 0.0613 mol of CaCN2 × (2 mol NH3 / 1 mol CaCN2) ≈ 0.1226 mol of NH3. Now, let's convert moles of NH3 to grams using the molar mass of NH3: Molar mass of NH3: N (14.01 g/mol) + 3H (3.03 g/mol) = 17.04 g/mol Grams of NH3: 0.1226 mol × 17.04 g/mol ≈ 2.09 g (theoretical yield)

Now we will use the given 86.0% yield to calculate the actual amount of NH3 produced: Actual yield of NH3: 2.09 g × 86.0% = 2.09 g × 0.86 ≈ 1.80 g Therefore, 1.80 grams of NH3 would be produced in the reaction with an 86.0% yield.