Math

Question4) How much heat is absorbed when 30.00 g of C(s)\mathrm{C}(\mathrm{s}) reacts in the presence of excess SO2( g)\mathrm{SO}_{2}(\mathrm{~g}) to produce CS2(l)\mathrm{CS}_{2}(l) and CO(g)\mathrm{CO}(g) according to the following chemical equation? 5C(s)+2SO2(g)CS2(l)+4CO(g)ΔH=+239.9 kJ5 \mathrm{C}(s)+2 \mathrm{SO}_{2}(g) \rightarrow \mathrm{CS}_{2}(l)+4 \mathrm{CO}(g) \quad \Delta H^{\circ}=+239.9 \mathrm{~kJ} A) 239.9 kJ B) 119.9 kJ C) 599.2 kJ D) 1439 kJ

Studdy Solution

STEP 1

1. The reaction is carried out under standard conditions.
2. The given enthalpy change (ΔH\Delta H^{\circ}) is for the reaction as written.
3. The molar mass of carbon (C\mathrm{C}) is approximately 12.01g/mol12.01 \, \text{g/mol}.

STEP 2

1. Calculate the number of moles of carbon (C\mathrm{C}) in 30.00 g.
2. Determine the fraction of the reaction that occurs with 30.00 g of carbon.
3. Calculate the heat absorbed based on the fraction of the reaction.

STEP 3

Calculate the number of moles of carbon (C\mathrm{C}) in 30.00 g:
Moles of C=mass of Cmolar mass of C \text{Moles of C} = \frac{\text{mass of C}}{\text{molar mass of C}} =30.00g12.01g/mol = \frac{30.00 \, \text{g}}{12.01 \, \text{g/mol}}
2.498mol \approx 2.498 \, \text{mol}

STEP 4

Determine the fraction of the reaction that occurs with 2.498 moles of carbon:
The balanced chemical equation shows that 5 moles of C\mathrm{C} are required for the reaction as written.
Fraction of reaction=moles of C5moles \text{Fraction of reaction} = \frac{\text{moles of C}}{5 \, \text{moles}} =2.498mol5mol = \frac{2.498 \, \text{mol}}{5 \, \text{mol}}
0.4996 \approx 0.4996

STEP 5

Calculate the heat absorbed based on the fraction of the reaction:
Heat absorbed=Fraction of reaction×ΔH \text{Heat absorbed} = \text{Fraction of reaction} \times \Delta H^{\circ} =0.4996×239.9kJ = 0.4996 \times 239.9 \, \text{kJ}
119.8kJ \approx 119.8 \, \text{kJ}
Since the closest option is 119.9 kJ, the heat absorbed is approximately:
119.9kJ \boxed{119.9 \, \text{kJ}}

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