Question

question 23
consider the following reaction;
mg + f₂ → mgf₂
how many grams of magnesium fluoride can be produced if you begin with 13.58 grams of each reactant?
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question 24
consider the following reaction:
2na + 2hcl → 2nacl + h₂
how many mols of hydrogen gas (h₂) can be produced if you begin with 23.55 grams of each reactant?
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question 25
consider the following reaction:
2hcl + caco₃ → cacl₂ + h₂o + co₂
how many mols of calcium chloride can be produced if you begin with 10.6 ml of 0.62 m hcl and 6.16 grams of calcium carbonate? record your answer in scientific notation, using 3 significant figures.
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Explanation:

Question 23

Step1: Find moles of each reactant

Molar mass of \( \text{Mg} = 24.305 \, \text{g/mol} \), moles of \( \text{Mg} = \frac{13.58 \, \text{g}}{24.305 \, \text{g/mol}} \approx 0.5587 \, \text{mol} \)
Molar mass of \( \text{F}_2 = 38.00 \, \text{g/mol} \), moles of \( \text{F}_2 = \frac{13.58 \, \text{g}}{38.00 \, \text{g/mol}} \approx 0.3574 \, \text{mol} \)

Step2: Determine limiting reactant

From reaction \( \text{Mg} + \text{F}_2
ightarrow \text{MgF}_2 \), 1:1 ratio.
\( \text{F}_2 \) is limiting (0.3574 mol < 0.5587 mol).

Step3: Calculate moles of \( \text{MgF}_2 \)

Moles of \( \text{MgF}_2 = \) moles of \( \text{F}_2 = 0.3574 \, \text{mol} \)

Step4: Convert to grams

Molar mass of \( \text{MgF}_2 = 24.305 + 2(18.998) = 62.301 \, \text{g/mol} \)
Mass of \( \text{MgF}_2 = 0.3574 \, \text{mol} \times 62.301 \, \text{g/mol} \approx 22.27 \, \text{g} \)

Step1: Find moles of each reactant

Molar mass of \( \text{Na} = 22.99 \, \text{g/mol} \), moles of \( \text{Na} = \frac{23.55 \, \text{g}}{22.99 \, \text{g/mol}} \approx 1.024 \, \text{mol} \)
Molar mass of \( \text{HCl} = 36.46 \, \text{g/mol} \), moles of \( \text{HCl} = \frac{23.55 \, \text{g}}{36.46 \, \text{g/mol}} \approx 0.6459 \, \text{mol} \)

Step2: Determine limiting reactant

Reaction: \( 2\text{Na} + 2\text{HCl}
ightarrow 2\text{NaCl} + \text{H}_2 \)
Moles of \( \text{Na} \) needed for \( \text{HCl} \): \( 0.6459 \, \text{mol (HCl)} \times \frac{2 \, \text{mol Na}}{2 \, \text{mol HCl}} = 0.6459 \, \text{mol Na} \)
\( \text{HCl} \) is limiting (needs 0.6459 mol Na, but we have 1.024 mol Na).

Step3: Calculate moles of \( \text{H}_2 \)

From reaction, \( 2 \, \text{mol HCl}
ightarrow 1 \, \text{mol H}_2 \)
Moles of \( \text{H}_2 = 0.6459 \, \text{mol HCl} \times \frac{1 \, \text{mol H}_2}{2 \, \text{mol HCl}} \approx 0.323 \, \text{mol} \)

Step1: Moles of \( \text{HCl} \) from solution

Molarity \( M = \frac{\text{moles}}{\text{volume (L)}} \), volume = \( 10.6 \, \text{mL} = 0.0106 \, \text{L} \)
Moles of \( \text{HCl} = 0.62 \, \text{M} \times 0.0106 \, \text{L} = 0.006572 \, \text{mol} \)

Step2: Moles of \( \text{CaCO}_3 \)

Molar mass of \( \text{CaCO}_3 = 100.09 \, \text{g/mol} \), moles of \( \text{CaCO}_3 = \frac{6.16 \, \text{g}}{100.09 \, \text{g/mol}} \approx 0.06155 \, \text{mol} \)

Step3: Determine limiting reactant

Reaction: \( 2\text{HCl} + \text{CaCO}_3
ightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2 \)
Moles of \( \text{HCl} \) needed for \( \text{CaCO}_3 \): \( 0.06155 \, \text{mol} \times 2 = 0.1231 \, \text{mol} \) (more than 0.006572 mol)
\( \text{HCl} \) is limiting.

Step4: Moles of \( \text{CaCl}_2 \)

From reaction, \( 2 \, \text{mol HCl}
ightarrow 1 \, \text{mol CaCl}_2 \)
Moles of \( \text{CaCl}_2 = \frac{0.006572 \, \text{mol HCl}}{2} = 0.003286 \, \text{mol} \)
In scientific notation: \( 3.29 \times 10^{-3} \, \text{mol} \) (3 sig figs)

Answer:

\( \approx 22.3 \, \text{g} \) (3 sig figs)

Question 24