Which of the following is/are correct?The following reaction occurs:CS2+3Cl2⟶ΔCCl4+S2Cl21.0 g of CS2 and 2.0 g of Cl2 reacts.

(Mw of CS2 = 7 6, Mw of Cl2 = 71, Mw of CCl4= 154 g mol-1) Weight of Cl2 needed =1.0g CS21 mol CS276 g CS23 mol CS2mol CS271 g Cl2mol Cl2=1×3×7176=2.8 g Cl2 needed Since there is 2.0 g Cl2 present, Cl2 is the limiting quantity 1. Weight of Cs2 used=2.0 g Cl21 mol Cl271 g Cl21 mol CS23 mol Cl276 g CS2mol CS2=2×1×1×7671×3=0.714 g CS2 used.2. Weight of CS2 excess or formed =(1.0 g CS2 present) - (0.714 g used) =0.286 g CS2 formed 3. Weight of CCl4 formed =(2.0 g Cl2) 1 mol Cl271 g Cl21 mol CCl43 mol Cl2 154 g CCl4mol CCl4=2×1×15471×3=1.45 g CCl4 4. Wrong

Which of the following is/are correct?The following reaction occurs:CS2+3Cl2⟶ΔCCl4+S2Cl21.0 g of CS2 and 2.0 g of Cl2 reacts.

(Mw of CS2 = 7 6, Mw of Cl2 = 71, Mw of CCl4= 154 g mol-1) Weight of Cl2 needed =1.0g CS21 mol CS276 g CS23 mol CS2mol CS271 g Cl2mol Cl2=1×3×7176=2.8 g Cl2 needed Since there is 2.0 g Cl2 present, Cl2 is the limiting quantity 1. Weight of Cs2 used=2.0 g Cl21 mol Cl271 g Cl21 mol CS23 mol Cl276 g CS2mol CS2=2×1×1×7671×3=0.714 g CS2 used.2. Weight of CS2 excess or formed =(1.0 g CS2 present) - (0.714 g used) =0.286 g CS2 formed 3. Weight of CCl4 formed =(2.0 g Cl2) 1 mol Cl271 g Cl21 mol CCl43 mol Cl2 154 g CCl4mol CCl4=2×1×15471×3=1.45 g CCl4 4. Wrong

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Which of the following is/are correct?
The following reaction occurs:
${CS}_{2} + 3 {Cl}_{2} \overset{Δ}{⟶} {CCl}_{4} + S_{2} {Cl}_{2}$
1.0 g of CS₂ and 2.0 g of Cl₂ reacts.

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0.8 g Cl₂ is in excess.

0.714 g CS₂ is used in the reaction.

1.45 g of CCl₄ is formed.

0.286 g CS₂ is in excess.

answer is A, B, C.

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Detailed Solution

$(Mw of {CS}_{2} = 7 6, Mw of {Cl}_{2} = 71, Mw of {CCl}_{4} = 154 g {mol}^{- 1}) Weight of {Cl}_{2} needed \begin{array}{l} = (1 .0 g {CS}_{2}) (\frac{1 mol {CS}_{2}}{76 g {CS}_{2}}) (\frac{3 mol {CS}_{2}}{mol {CS}_{2}}) (\frac{71 g {Cl}_{2}}{mol {Cl}_{2}}) \\ = \frac{1 \times 3 \times 71}{76} = 2 .8 g {Cl}_{2} needed \\ Since there is 2.0 g {Cl}_{2} present, {Cl}_{2} is the limiting quantity \\ 1 . Weight of {Cs}_{2} used \\ = (2 .0 g {Cl}_{2}) (\frac{1 mol {Cl}_{2}}{71 g {Cl}_{2}}) (\frac{1 mol {CS}_{2}}{3 mol {Cl}_{2}}) (\frac{76 g {CS}_{2}}{mol {CS}_{2}}) \\ = \frac{2 \times 1 \times 1 \times 76}{71 \times 3} = 0.714 g {CS}_{2} used . \end{array}$

$2 . Weight of {CS}_{2} excess or formed = (1.0 g {CS}_{2} present) - (0.714 g used) = 0.286 g {CS}_{2} formed 3 . Weight of {CCl}_{4} formed = (2.0 g {Cl}_{2}) (\frac{1 mol {Cl}_{2}}{71 g {Cl}_{2}}) (\frac{1 mol {CCl}_{4}}{3 mol {Cl}_{2}}) (\frac{154 g {CCl}_{4}}{mol {CCl}_{4}}) = \frac{2 \times 1 \times 154}{71 \times 3} = 1 .45 g {CCl}_{4} 4 . Wrong$