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Liula [17]
3 years ago
14

Methane (CH4) reacts with Cl2 to yield CCl4 and HCl by the following reaction equation: CH4 + 4 Cl2 → CCl4 + 4HCl. What is the Δ

H of the reaction if 51.3 g of CH4 reacts with excess Cl2 to yield 1387.6 kJ?
Chemistry
1 answer:
Vlad1618 [11]3 years ago
4 0

Answer: The ΔH of the reaction if 51.3 g of CH_4 reacts with excess Cl_2 to yield 1387.6 kJ is 432.27kJ

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

moles of CH_4

\text{Number of moles of methane}=\frac{51.3g}{16g/mol}=3.21moles

CH_4+4Cl_2\rightarrow CCl_4+4HCl   \Delta H=?

As Cl_2 is present in excess, CH_4 is the limiting reagent as it limits the formation of product.

If 3.21 moles of methane releases heat = 1387.6 kJ

Thus 1 mole of methane release=\frac{1387.6}{3.21}\times 1=432.27kJ

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(b). Mass and distance.

Explanation:

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7 0
3 years ago
When 150. g zinc sulfide are burned in excess oxygen, 68.5 g of zinc oxide are actually produced, along with sulfur dioxide. Det
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%yield = 54.6%

<h3>Further explanation</h3>

Percent yield is the compare of the amount of product obtained from a reaction with the amount you calculated

(theoretical)

General formula:

Percent yield = (Actual yield / theoretical yield )x 100%

<h3 />

Reaction

2ZnS+3O₂ ⇒ 2ZnO+2SO₂

MW ZnS = 97.474 g/mol

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\tt \dfrac{150}{97.474}=1.54

MW ZnO = 81.38 g/mol

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4 0
3 years ago
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