Question

Calcium cyanamide reacts with water to form calcium carbonate and ammonia. $$ \mathrm{CaCN}_{2}(s)+3 \mathrm{H}_{2} \mathrm{O}(I) \longrightarrow \mathrm{CaCO}_{3}(s)+2 \mathrm{NH}_{3}(g) $$ a. How many grams of water are needed to react with \(75.0 \mathrm{~g}\) of \(\mathrm{CaCN}_{2}\) ? b. How many grams of \(\mathrm{NH}_{3}\) are produced from \(5.24 \mathrm{~g}\) of \(\mathrm{CaCN}_{2} ?\) c. How many grams of \(\mathrm{CaCO}_{3}\) form if \(155 \mathrm{~g}\) of water reacts?

Short answer

a) 50.6 g of water. b) 2.23 g of NH3. c) 287 g of CaCO3.

Step-by-step solution

Write and balance the equation

The given chemical reaction is already balanced: \[ \mathrm{CaCN}_{2}(s) + 3 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{CaCO}_{3}(s) + 2 \mathrm{NH}_{3}(g) \]

Molar masses calculation

Calculate the molar masses of the reactants and products: \[ M(\mathrm{CaCN}_{2}) = 40.08 + 2(12.01 + 14.01) = 80.10 \text{ g/mol} \]\[ M(\mathrm{H}_{2} \mathrm{O}) = 2(1.01) + 16.00 = 18.02 \text{ g/mol} \]\[ M(\mathrm{NH}_{3}) = 14.01 + 3(1.01) = 17.04 \text{ g/mol} \]\[ M(\mathrm{CaCO}_{3}) = 40.08 + 12.01 + 3(16.00) = 100.09 \text{ g/mol} \]

Step 3a: Calculate moles of \(\mathrm{CaCN}_{2}\) for part a

First, calculate the number of moles of \(\mathrm{CaCN}_{2}\): \[ n(\mathrm{CaCN}_{2}) = \frac{75.0 \text{ g}}{80.10 \text{ g/mol}} = 0.936 \text{ mol} \]

Step 3b: Calculate moles of \(\mathrm{H}_{2} \mathrm{O}\) for part a

According to the balanced equation, 1 mole of \(\mathrm{CaCN}_{2}\) reacts with 3 moles of \(\mathrm{H}_{2} \mathrm{O}\). So, \[ n(\mathrm{H}_{2} \mathrm{O}) = 3 \times n(\mathrm{CaCN}_{2}) = 3 \times 0.936 = 2.808 \text{ mol} \]

Step 3c: Calculate mass of \(\mathrm{H}_{2} \mathrm{O}\) for part a

Calculate the mass of \(\mathrm{H}_{2} \mathrm{O}\) needed: \[ m(\mathrm{H}_{2} \mathrm{O}) = n(\mathrm{H}_{2} \mathrm{O}) \times M(\mathrm{H}_{2} \mathrm{O}) = 2.808 \text{ mol} \times 18.02 \text{ g/mol} = 50.6 \text{ g} \]

Step 4a: Calculate moles of \(\mathrm{CaCN}_{2}\) for part b

First, calculate the number of moles of \(\mathrm{CaCN}_{2}\) used: \[ n(\mathrm{CaCN}_{2}) = \frac{5.24 \text{ g}}{80.10 \text{ g/mol}} = 0.0654 \text{ mol} \]

Step 4b: Calculate moles of \(\mathrm{NH}_{3}\) for part b

According to the balanced equation, 1 mole of \(\mathrm{CaCN}_{2}\) produces 2 moles of \(\mathrm{NH}_{3}\). So, \[ n(\mathrm{NH}_{3}) = 2 \times n(\mathrm{CaCN}_{2}) = 2 \times 0.0654 = 0.131 \text{ mol} \]

Step 4c: Calculate mass of \(\mathrm{NH}_{3}\) for part b

Calculate the mass of \(\mathrm{NH}_{3}\) produced: \[ m(\mathrm{NH}_{3}) = n(\mathrm{NH}_{3}) \times M(\mathrm{NH}_{3}) = 0.131 \text{ mol} \times 17.04 \text{ g/mol} = 2.23 \text{ g} \]

Step 5a: Calculate moles of \(\mathrm{H}_{2} \mathrm{O}\) for part c

First, calculate the number of moles of \(\mathrm{H}_{2} \mathrm{O}\) used: \[ n(\mathrm{H}_{2} \mathrm{O}) = \frac{155 \text{ g}}{18.02 \text{ g/mol}} = 8.60 \text{ mol} \]

Step 5b: Calculate moles of \(\mathrm{CaCN}_{2}\) that reacts with \(\mathrm{H}_{2} \mathrm{O}\) for part c

According to the balanced equation, 3 moles of \(\mathrm{H}_{2} \mathrm{O}\) react with 1 mole of \(\mathrm{CaCN}_{2}\). So, \[ n(\mathrm{CaCN}_{2}) = \frac{n(\mathrm{H}_{2} \mathrm{O})}{3} = \frac{8.60}{3} = 2.87 \text{ mol} \]

Step 5c: Calculate mass of \(\mathrm{CaCO}_{3}\) for part c

According to the balanced equation, 1 mole of \(\mathrm{CaCN}_{2}\) produces 1 mole of \(\mathrm{CaCO}_{3}\). So, \[ m(\mathrm{CaCO}_{3}) = n(\mathrm{CaCN}_{2}) \times M(\mathrm{CaCO}_{3}) = 2.87 \text{ mol} \times 100.09 \text{ g/mol} = 287 \text{ g} \]

Key concepts

Molar Mass Calculation

To understand any chemical reaction, we first need to know the molar mass of each compound involved. The molar mass of a substance is the mass of one mole of that substance and is usually expressed in grams per mole (g/mol). You can find this by adding up the atomic masses of all atoms in a molecule. For example, in the chemical reaction given:

• Molar mass of \(\text{CaCN}_2\) = 40.08 (for Ca) + 2 * (12.01 (for each C) + 14.01 (for each N)) = 80.10 g/mol
• Molar mass of \(\text{H}_2\text{O}\) = 2 * 1.01 (for each H) + 16.00 (for O) = 18.02 g/mol
• Molar mass of \(\text{NH}_3\) = 14.01 (for N) + 3 * 1.01 (for each H) = 17.04 g/mol
• Molar mass of \(\text{CaCO}_3\) = 40.08 (for Ca) + 12.01 (for C) + 3 * 16.00 (for each O) = 100.09 g/mol

Understanding molar mass helps us in calculations related to converting mass to moles and vice versa.

Mole-to-Mole Ratio

In a balanced chemical equation, the coefficients indicate the mole-to-mole ratio. This ratio tells us how many moles of one substance will react with another. For example, the balanced equation \(\text{CaCN}_2 + 3\text{H}_2\text{O} \rightarrow \text{CaCO}_3 + 2\text{NH}_3\) shows:

• 1 mole of \(\text{CaCN}_2\) reacts with 3 moles of \(\text{H}_2\text{O}\)
• 1 mole of \(\text{CaCN}_2\) produces 1 mole of \(\text{CaCO}_3\)
• 1 mole of \(\text{CaCN}_2\) produces 2 moles of \(\text{NH}_3\)

Knowing these ratios allows us to determine how much of each product forms or how much of each reactant is needed in a given reaction.

Mass-Mass Calculations

Mass-mass calculations involve converting the mass of a reactant to the mass of a product using their respective mole-to-mole ratios and molar masses. Here’s a step-by-step approach:

• First, convert the mass of the reactant to moles by dividing by its molar mass: \[ n(\text{CaCN}_2) = \frac{75.0 \text{ g}}{80.10 \text{ g/mol}} = 0.936 \text{ mol} \]
• Next, use the mole-to-mole ratio to convert moles of the reactant to moles of the product: \[ n(\text{H}_2\text{O}) = 3 \times n(\text{CaCN}_2) = 3 \times 0.936 = 2.808 \text{ mol} \]
• Finally, convert the moles of the product back to mass using its molar mass: \[ m(\text{H}_2\text{O}) = n(\text{H}_2\text{O}) \times M(\text{H}_2\text{O}) = 2.808 \text{ mol} \times 18.02 \text{ g/mol} = 50.6 \text{ g} \]

This method can be adapted for any reactant-product pair in a balanced equation.

Balanced Chemical Equation

A balanced chemical equation is crucial for stoichiometric calculations. It ensures that the number of atoms of each element is the same on both sides of the equation, satisfying the Law of Conservation of Mass. In the equation \(\text{CaCN}_2 + 3 \text{H}_2\text{O} \rightarrow \text{CaCO}_3 + 2 \text{NH}_3\),:

• The left side has: 1 Ca, 2 C, 2 N, and 3 * (2 H + 1 O)
• The right side has: 1 Ca, 1 C (in \(\text{CaCO}_3\) + 1 C in \(\text{CO}_3\), 2 N (in \(\text{NH}_3\)), and 3 * O (in \(\text{CaCO}_3\)), and 2 * 3 H (in \(\text{NH}_3\))

Balancing helps us understand the exact stoichiometric relationships between reactants and products, making sure we use and produce the right amounts of substances.