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3 years ago

2 Al + 3 H2SO4 --> Al2(SO4)3 + 3 H2

Chemistry

1 answer:

suter [353]3 years ago

3 0

Answer:

The answer to your question is 0.15 g of H₂

Explanation:

Data

mass of H₂ =

mass of H₂SO₄ = 7.5 g

Balanced chemical reaction

2Al + 3H₃SO₄ ⇒ Al₂(SO₄)₃ + 3H₂

Process

1.- Calculate the molar mass of hydrogen and sulfuric acid

H₂ = 3(2) = 6 g

H₂SO₄ = 2 + 32 + (16 x 4) = 3 x 98 g = 294 g

2.- Use proportions to solve this problem

294 g of H₂SO₄ ---------------- 6 g of H₂

7.5 g of H₂SO₄ ---------------- x

x = (7.5 x 6)/294

x = 45/294

x = 0.15 g of H₂

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Answer:

The enthalpy change during the reaction is -199. kJ/mol.

Explanation:

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First we have to calculate the heat gained by the solution in coffee-cup calorimeter.

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c = specific heat = $4.18 J/^oC$

$T_{final}$ = final temperature = $23.1^oC$

$T_{initial}$ = initial temperature = $28.9^oC$

Now put all the given values in the above formula, we get:

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Now we have to calculate the enthalpy change during the reaction.

$\Delta H=-\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat gained = 2.242 kJ

n = number of moles fructose = $\frac{\text{Mass of manganese}}{\text{Molar mass of manganese}}=\frac{0.620 g}{54.94 g/mol}=0.0113 mol$

$\Delta H=-\frac{2.242 kJ}{0.0113 mol }=-199. kJ/mol$

Therefore, the enthalpy change during the reaction is -199. kJ/mol.

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