Question

question 26
in a process to vulcanize rubber, a process that makes rubber stronger, disulfur dichloride is created by chlorine gas reacting with molten sulfur in the following reaction:
$s_8 + 4cl_2 \
ightarrow 4s_2cl_2$
if 2.43 mol of sulfur reacts with 5.01 mol of chlorine...
which is the limiting reactant:
$\bigcirc cl_2$
$\bigcirc s_8$
how many moles of disulfur dichloride is produced?
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question 27
given the following equation:
$ca(oh)_2 + 2 hcl \
ightarrow cacl_2 + 2 h_2o$
how many grams of $cacl_2$ are produced when 89.2 grams of $ca(oh)_2$ reacts with 96.4 grams of hcl?
$\square$ g $cacl_2$
what is the limiting reactant?

Explanation:

Question 26 - Limiting Reactant and Moles of Product

Step1: Determine Limiting Reactant

From the reaction \( S_8 + 4Cl_2
ightarrow 4S_2Cl_2 \), the mole ratio of \( S_8 \) to \( Cl_2 \) is \( 1:4 \).

• Moles of \( S_8 = 2.43 \, \text{mol} \)
• Moles of \( Cl_2 = 5.01 \, \text{mol} \)

Calculate moles of \( Cl_2 \) required to react with \( 2.43 \, \text{mol} \) of \( S_8 \):
\( \text{Required } Cl_2 = 2.43 \, \text{mol} \, S_8 \times \frac{4 \, \text{mol} \, Cl_2}{1 \, \text{mol} \, S_8} = 9.72 \, \text{mol} \, Cl_2 \)

We have only \( 5.01 \, \text{mol} \) of \( Cl_2 \), which is less than \( 9.72 \, \text{mol} \). So \( Cl_2 \) is the limiting reactant.

Step2: Calculate Moles of \( S_2Cl_2 \)

From the reaction, mole ratio of \( Cl_2 \) to \( S_2Cl_2 \) is \( 4:4 = 1:1 \).
Moles of \( S_2Cl_2 = \text{Moles of limiting reactant } Cl_2 = 5.01 \, \text{mol} \) (since \( 4 \, \text{mol} \, Cl_2 \) produces \( 4 \, \text{mol} \, S_2Cl_2 \), so moles of \( S_2Cl_2 \) equal moles of \( Cl_2 \) used).

Question 27 - Grams of Product and Limiting Reactant

Step1: Molar Masses

• \( \text{Molar mass of } Ca(OH)_2 = 40.08 + 2(16.00 + 1.01) = 74.10 \, \text{g/mol} \)
• \( \text{Molar mass of } HCl = 1.01 + 35.45 = 36.46 \, \text{g/mol} \)
• \( \text{Molar mass of } CaCl_2 = 40.08 + 2(35.45) = 110.98 \, \text{g/mol} \)

Step2: Moles of Reactants

• Moles of \( Ca(OH)_2 = \frac{89.2 \, \text{g}}{74.10 \, \text{g/mol}} \approx 1.204 \, \text{mol} \)
• Moles of \( HCl = \frac{96.4 \, \text{g}}{36.46 \, \text{g/mol}} \approx 2.644 \, \text{mol} \)

Step3: Limiting Reactant (Mole Ratio)

Reaction: \( Ca(OH)_2 + 2HCl
ightarrow CaCl_2 + 2H_2O \), mole ratio \( Ca(OH)_2:HCl = 1:2 \).

• Moles of \( HCl \) required for \( 1.204 \, \text{mol} \, Ca(OH)_2 \): \( 1.204 \times 2 = 2.408 \, \text{mol} \)
• We have \( 2.644 \, \text{mol} \, HCl \), which is more than \( 2.408 \, \text{mol} \). Thus, \( Ca(OH)_2 \) is the limiting reactant.

Step4: Grams of \( CaCl_2 \)

Mole ratio \( Ca(OH)_2:CaCl_2 = 1:1 \), so moles of \( CaCl_2 = 1.204 \, \text{mol} \).
Grams of \( CaCl_2 = 1.204 \, \text{mol} \times 110.98 \, \text{g/mol} \approx 133.6 \, \text{g} \)

Answer:

Question 26:

• Limiting Reactant: \( \boldsymbol{Cl_2} \)
• Moles of \( S_2Cl_2 \): \( \boldsymbol{5.01 \, \text{mol}} \)

Question 27:

• Grams of \( CaCl_2 \): \( \boldsymbol{\approx 133.6 \, \text{g}} \)
• Limiting Reactant: \( \boldsymbol{Ca(OH)_2} \)