Question

1. brodbelt 3200428 10 pts possible

consider the following reaction:
\ce{cacn2 + 3 h2o -> caco3 + 2 nh3}
105.0 g \ce{cacn2} and 78.0 g \ce{h2o} are reacted. assuming 100% efficiency, which reactant is in excess and how much is leftover? the molar mass of \ce{cacn2} is 80.11 g/mol. the molar mass of \ce{caco3} is 100.09 g/mol.
\\(\boldsymbol{\circ}\\) 1. \ce{h2o}; 10.7 g left over
\\(\boldsymbol{\circ}\\) 2. \ce{h2o}; 70.8 g left over
\\(\boldsymbol{\circ}\\) 3. \ce{cacn2}; 10.7 g left over
\\(\boldsymbol{\circ}\\) 4. \ce{cacn2}; 7.20 g left over
\\(\boldsymbol{\circ}\\) 5. \ce{cacn2}; 70.8 g left over
\\(\boldsymbol{\circ}\\) 6. \ce{h2o}; 7.20 g left over

Explanation:

Step 1: Calculate moles of \( \text{CaCN}_2 \)

Moles of \( \text{CaCN}_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{105.0\ \text{g}}{80.11\ \text{g/mol}} \approx 1.311\ \text{mol} \)

Step 2: Calculate moles of \( \text{H}_2\text{O} \)

Molar mass of \( \text{H}_2\text{O} = 18.02\ \text{g/mol} \) (since \( 2(1.008) + 16.00 = 18.016 \approx 18.02 \))
Moles of \( \text{H}_2\text{O} = \frac{78.0\ \text{g}}{18.02\ \text{g/mol}} \approx 4.328\ \text{mol} \)

Step 3: Determine stoichiometric ratio

From the reaction \( \text{CaCN}_2 + 3\text{H}_2\text{O} ightarrow \text{CaCO}_3 + 2\text{NH}_3 \), the ratio of \( \text{CaCN}_2 \) to \( \text{H}_2\text{O} \) is \( 1:3 \).

For \( 1.311\ \text{mol}\ \text{CaCN}_2 \), moles of \( \text{H}_2\text{O} \) required = \( 1.311\ \text{mol} \times 3 = 3.933\ \text{mol} \)

Step 4: Identify limiting and excess reactant

We have \( 4.328\ \text{mol}\ \text{H}_2\text{O} \), which is more than \( 3.933\ \text{mol} \) required. So \( \text{H}_2\text{O} \) is in excess? Wait, no—wait, let's check the other way. Wait, maybe I mixed up. Wait, let's check how much \( \text{CaCN}_2 \) is needed for \( \text{H}_2\text{O} \).

Moles of \( \text{CaCN}_2 \) required for \( 4.328\ \text{mol}\ \text{H}_2\text{O} = \frac{4.328\ \text{mol}}{3} \approx 1.443\ \text{mol} \). But we have only \( 1.311\ \text{mol}\ \text{CaCN}_2 \), so \( \text{CaCN}_2 \) is limiting, and \( \text{H}_2\text{O} \) is excess? Wait, no—wait, no, let's re-express.

Wait, the reaction is 1 mol \( \text{CaCN}_2 \) reacts with 3 mol \( \text{H}_2\text{O} \).

If \( \text{CaCN}_2 \) is 1.311 mol, then \( \text{H}_2\text{O} \) needed is \( 1.311 \times 3 = 3.933 \) mol. We have 4.328 mol \( \text{H}_2\text{O} \), so excess \( \text{H}_2\text{O} \) moles = \( 4.328 - 3.933 = 0.395\ \text{mol} \)

Wait, but the options have \( \text{CaCN}_2 \) as excess? Wait, maybe I made a mistake. Wait, let's recalculate.

Wait, maybe I flipped. Let's check the mass of \( \text{CaCN}_2 \) and \( \text{H}_2\text{O} \) again. Wait, the problem says 105.0 g \( \text{CaCN}_2 \) and 78.0 g \( \text{H}_2\text{O} \).

Wait, let's calculate the mass of \( \text{H}_2\text{O} \) required to react with 105.0 g \( \text{CaCN}_2 \).

Moles of \( \text{CaCN}_2 = 105.0 / 80.11 ≈ 1.3107\ \text{mol} \)

Moles of \( \text{H}_2\text{O} \) required = \( 1.3107 \times 3 = 3.9321\ \text{mol} \)

Mass of \( \text{H}_2\text{O} \) required = \( 3.9321\ \text{mol} \times 18.02\ \text{g/mol} ≈ 70.86\ \text{g} \)

We have 78.0 g \( \text{H}_2\text{O} \), so excess \( \text{H}_2\text{O} \) = \( 78.0 - 70.86 ≈ 7.14\ \text{g} \)? Wait, but the options have \( \text{CaCN}_2 \) as excess? Wait, no—wait, maybe I messed up the limiting reactant. Wait, let's check the other way: how much \( \text{CaCN}_2 \) is needed for 78.0 g \( \text{H}_2\text{O} \).

Moles of \( \text{H}_2\text{O} = 78.0 / 18.02 ≈ 4.328\ \text{mol} \)

Moles of \( \text{CaCN}_2 \) required = \( 4.328 / 3 ≈ 1.4427\ \text{mol} \)

Mass of \( \text{CaCN}_2 \) required = \( 1.4427\ \text{mol} \times 80.11\ \text{g/mol} ≈ 115.5\ \text{g} \)

But we have only 105.0 g \( \text{CaCN}_2 \), which is less than 115.5 g. So \( \text{CaCN}_2 \) is limiting, and \( \text{H}_2\text{O} \) is in excess. Wait, but the options have \( \text{CaCN}_2 \) as excess? Wait, no—wait, the options are:

1. \( \text{H}_2\text{O} \); 10.7 g left over
2. \( \text{H}_2\text{O} \); 70.8 g left over
3. \( \text{CaCN}_2 \); 10.7 g left over
4. \( \text{CaCN}_2 \); 7.20 g left over
5. \( \text{CaCN}_2 \); 70.8 g left over
6. \( \text{H}_2\text{O} \); 7.20 g left over

Wait, maybe I made a mistake. Wait, let's recalculate the excess.

Wait, if \( \text{CaCN}_2 \) is limiting (1.3107 mol), then moles of \( \text{H}_2\text{O} \) used = 3.9321 mol. Mass of \( \text{H}_2\text{O} \) used = 3.9321 mol × 18.02 g/mol ≈ 70.86 g. So excess \( \text{H}_2\text{O} \) = 78.0 g - 70.86 g ≈ 7.14 g, which is approximately 7.20 g. Wait, but option 6 is \( \text{H}_2\text{O} \); 7.20 g left over. Wait, but earlier I thought \( \text{CaCN}_2 \) was limiting, but maybe I flipped. Wait, no—wait, let's check the problem again.

Wait, the reaction is \( \text{CaCN}_2 + 3\text{H}_2\text{O} ightarrow \text{CaCO}_3 + 2\text{NH}_3 \). So 1 mol \( \text{CaCN}_2 \) reacts with 3 mol \( \text{H}_2\text{O} \).

Given:
- \( \text{CaCN}_2 \): 105.0 g (molar mass 80.11 g/mol) → moles = 105.0 / 80.11 ≈ 1.3107 mol
- \( \text{H}_2\text{O} \): 78.0 g (molar mass 18.02 g/mol) → moles = 78.0 / 18.02 ≈ 4.328 mol

Stoichiometric ratio: \( \text{CaCN}_2 : \text{H}_2\text{O} = 1:3 \).

So, for \( \text{CaCN}_2 \) (1.3107 mol), \( \text{H}_2\text{O} \) needed is 1.3107 × 3 = 3.9321 mol. We have 4.328 mol \( \text{H}_2\text{O} \), so \( \text{H}_2\text{O} \) is in excess. The excess moles of \( \text{H}_2\text{O} \) is 4.328 - 3.9321 = 0.3959 mol. Mass of excess \( \text{H}_2\text{O} \) = 0.3959 mol × 18.02 g/mol ≈ 7.13 g, which is approximately 7.20 g. So option 6: \( \text{H}_2\text{O} \); 7.20 g left over. Wait, but the options have option 6 as \( \text{H}_2\text{O} \); 7.20 g left over. But wait, maybe I made a mistake in molar mass of \( \text{H}_2\text{O} \). Wait, 18.015 g/mol is more accurate. Let's recalculate with 18.015 g/mol.

Moles of \( \text{H}_2\text{O} \) = 78.0 / 18.015 ≈ 4.330 mol.

Moles of \( \text{H}_2\text{O} \) used = 3.9321 mol (from \( \text{CaCN}_2 \) moles).

Excess moles \( \text{H}_2\text{O} \) = 4.330 - 3.9321 = 0.3979 mol.

Mass of excess \( \text{H}_2\text{O} \) = 0.3979 mol × 18.015 g/mol ≈ 7.17 g, which is ~7.20 g. So option 6 is correct. Wait, but the options are:

6. \( \text{H}_2\text{O} \); 7.20 g left over

Yes, that matches. Wait, but earlier I thought maybe \( \text{CaCN}_2 \) was excess, but no—because \( \text{CaCN}_2 \) is limiting. Wait, no—wait, if \( \text{H}_2\text{O} \) is in excess, then the excess is \( \text{H}_2\text{O} \). So the correct answer is option 6.

Answer:

1. \( \text{H}_2\text{O} \); 7.20 g left over