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3 years ago

1 Srcl (aq) + 1 H,60,(aq) → 2 HCl(aq) + 1 Srso (s)

Chemistry

1 answer:

Mrac [35]3 years ago

3 0

Answer:

Mass = 245.72 g

Explanation:

Given data:

Mass of SrCl₂ react = ?

Mass of H₂SO₄ = 300.0 g

Solution:

SrCl₂ + 2H₂SO₄ → 2HCl + Sr(HSO₄)₂

Number of moles of H₂SO₄:

Number of moles = mass/molar mass

Number of moles = 300.0 g/ 98.079 g/mol

Number of moles = 3.1 mol

Now we will compare the moles of SrCl₂ and H₂SO₄.

H₂SO₄ : SrCl₂

2 : 1

3.1 : 1/2×3.1 = 1.55 mol

Mass of SrCl₂:

Mass = number of moles × molar mass

Mass = 1.55 mol × 158.53 g/mol

Mass = 245.72 g

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<u>Answer:</u> The amount of heat produced by the reaction is -21.36 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]$

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$\Delta H_{rxn}=[(2\times \Delta H_f_{(SO_3(g))})]-[(2\times \Delta H_f_{(SO_2(g))})+(1\times \Delta H_f_{(O_2(g))})]$

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To calculate the number of moles, we use ideal gas equation, which is:

$PV=nRT$

where,

P = pressure of the gas = 1.00 bar

V = Volume of the gas = 2.67 L

n = number of moles of gas = ?

R = Gas constant = $0.0831\text{ L. bar }mol^{-1}K^{-1}$

T = temperature of the mixture = $25^oC=[25+273]K=298K$

Putting values in above equation, we get:

$1.00bar\times 2.67L=n\times 0.0831\text{ L. bar }mol^{-1}K^{-1}\times 298K\\\\n=\frac{1\times 2.67}{0.0831\times 298}=0.108mol$

To calculate the heat released of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

q = amount of heat released = ?

n = number of moles = 0.108 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction = -197.8 kJ/mol

Putting values in above equation, we get:

$-197.8kJ/mol=\frac{q}{0.108mol}\\\\q=(-197.8kJ/mol\times 0.108mol)=-21.36kJ$

Hence, the amount of heat produced by the reaction is -21.36 kJ

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