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4 years ago

if you begin with 210.3 g of Cl2, how many grams of HCl will you end up with? CH4 + 3Cl2 → CHCl3 + 3HCl *

Chemistry

1 answer:

tatyana61 [14]4 years ago

7 0

Answer:

108.04g

Explanation:

Looking at the balanced chemical equation theoretically, we can see that 3 moles of chlorine yielded 3 moles of HCl. This means that they have equal mole ratio.

Now let’s get the actual reactive moles. The number of moles is obtained by dividing the mass by the molar mass.

The molecular mass of chlorine gas is (2 * 35.5) = 72g/mol

The number of moles is thus 210.3/71 = 2.96 moles

Since the number of moles are equal, the number of moles of HCl produced too is 2.96 moles.

Now to get the mass of HCl produced, we multiply the number of moles by the molar mass of HCl. The molar mass of HCl is 36.5g/mol

The mass is thus 2.96 * 36.5 = 108.04g

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An aqueous potassium carbonate solution is made by dissolving 5.51 moles of K 2 CO 3 in sufficient water so that the final volum

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Answer:

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Explanation:

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3 years ago

Read 2 more answers

The question is in the picture below

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Answer:

$\Delta\text{H}_1+2\Delta\text{H}_2-\Delta\text{H}_3$

Explanation:

Hess's Law of Constant Heat Summation states that if a chemical equation can be written as the sum of several other chemical equations, the enthalpy change of the first chemical equation is equal to the sum of the enthalpy changes of the other chemical equations. Thus, the reaction that involves the conversion of reactant A to B, for example, has the same enthalpy change even if you convert A to C, before converting it to B. Regardless of how many steps it takes for the reactant to be converted to the product, the enthalpy change of the overall reaction is constant.

With Hess's Law in mind, let's see how A can be converted to 2C +E.

$\bf{\text{A} \rightarrow 2\text{B}}$ (Δ $\text{H}_1$ ) -----(1)

Since we have 2B, multiply the whole of II. by 2:

$\bf{2\text{B} \rightarrow 2\text{C} +2\text{D}}$ (2Δ $\text{H}_2$ ) -----(2)

This step converts all the B intermediates to 2C +2D. This means that the overall reaction at this stage is $\text{A} \rightarrow 2\text{C} +2\text{D}$ .

Reversing III. gives us a negative enthalpy change as such:

$\bf{2\text{D} \rightarrow \text{E}}$ (-Δ $\text{H}_3$ ) -----(3)

This step converts all the D intermediates formed from step (2) to E. This results in the overall equation of $\text{A} \rightarrow 2\text{C} +\text{E}$ , which is also the equation of interest.

Adding all three together:

$\text{A} \rightarrow 2\text{C}+\text{E}$ ( $\bf{\Delta\text{H}_1+2\Delta\text{H}_2-\Delta\text{H}_3 }$ )

Thus, the first option is the correct answer.

Supplementary:

To learn more about Hess's Law, do check out: brainly.com/question/26491956

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