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-BARSIC- [3]
3 years ago
12

The change in enthalpy is the change in heat energy of a reaction and is calculated by (Hint: Hess' Law): Group of answer choice

s
a. Sum of reactant enthalpies plus the sum of product enthalpies
b. Sum of reactant enthalpies minus the sum of product enthalpies
c. Sum of product enthalpies times the sum of reactant enthalpies
d. Sum of product enthalpies minus the sum of reactant enthalpies
Chemistry
1 answer:
Stolb23 [73]3 years ago
8 0

Answer:

d. Sum of product enthalpies minus the sum of reactant enthalpies

Explanation:

The standard enthalpy change of a reaction (ΔH°rxn) can be calculated using the following expression:

ΔH°rxn = ∑n(products) × ΔH°f(products) - ∑n(reactants) × ΔH°f(reactants)

where,

ni are the moles of products and reactants

ΔH°f(i) are the standard enthalpies of formation of products and reactants

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New questionWhat mass of calcium chloride (CaCl₂) would beproduced from the reaction of 125.9 g of hydrochloriacid (HCI) with ex
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Answer:

191.6 g of CaCl₂.

Explanation:

What is given?

Mass of HCl = 125.9 g.

Molar mass of CaCl₂ = 110.8 g/mol.

Molar mass of HCl = 36.4 g/mol.

Step-by-step solution:

First, we have to state the chemical equation. Ca(OH)₂ react with HCl to produce CaCl₂:

Ca(OH)_2+2HCl\rightarrow CaCl_2+2H_2O.

Now, let's convert 125.9 g of HCl to moles using the given molar mass (remember that the molar mass of a compound can be found using the periodic table). The conversion will look like this:

125.9\text{ g HCl}\cdot\frac{1\text{ mol HCl}}{36.4\text{ g HCl}}=3.459\text{ moles HCl.}

Let's find how many moles of CaCl₂ are being produced by 3.459 moles of HCl. You can see in the chemical equation that 2 moles of HCl reacted with excess Ca(OH)₂ produces 1 mol of CaCl₂, so we state a rule of three and the calculation is:

3.459\text{ moles HCl}\cdot\frac{1\text{ mol CaCl}_2}{2\text{ moles HCl}}=1.729\text{ moles CaCl}_2.

The final step is to find the mass of CaCl₂ using the molar mass of CaCl₂. This conversion will look like this:

1.729\text{ moles CaCl}_2\cdot\frac{110.8\text{ g CaCl}_2}{1\text{ mol CaCl}_2}=191.6\text{ g CaCl}_2.

The answer would be that we're producing a mass of 191.6 g of CaCl₂.

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