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valentina_108 [34]
2 years ago
11

2Al + 6HCl → 2AlCl3 + 3H2

Chemistry
1 answer:
Nadya [2.5K]2 years ago
6 0

Answer: The reaction produces 2.93 g H₂.

M_r:                        133.34  2.016

       2Al + 6HCl → 2AlCl₃ + 3H₂

Moles of AlCl₃ = 129 g AlCl₃ × (1 mol AlCl₃/133.34 g AlCl₃) = 0.9675 mol AlCl₃

Moles of H₂ = 0.9675 mol AlCl₃ × (3 mol H₂/2 mol AlCl₃) = 1.451 mol H₂

Mass of H₂ = 1.451 mol H₂ × (2.016 g H₂/1 mol H₂) = 2.93 g H₂

Explanation:

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<h3>How to determine the heat required change the temperature from –38 °C to 0 °C </h3>
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<h3>How to determine the heat required to melt the ice at 0 °C</h3>
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<h3>How to determine the heat required to change the temperature from 0 °C to 100 °C </h3>
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  • Initial temperature (T₁) = 0 °C
  • Final temperature (T₂) = 100 °C
  • Change in temperature (ΔT) = 100 – 0 = 100 °C
  • Specific heat capacity (C) = 4180 J/(kg·°C)
  • Heat (Q₃) =?

Q = MCΔT

Q₃ = 0.4 × 4180 × 100

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<h3>How to determine the heat required to vaporize the water at 100 °C</h3>
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Q₄ = 0.4 × 2260000

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Q₅ = 47904 J

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Qₜ = Q₁ + Q₂ + Q₃ + Q₄ + Q₅

Qₜ = 31160 + 133600 + 167200 + 904000 + 47904

Qₜ = 1.28×10⁶ J

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brainly.com/question/10286596

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