Question

An alloy of Aluminium, and Zinc was treated with aqueous HCI. The Aluminium and zinc dissolved according to the reaction:
2A1 - 6H2A1 + 3H,
Zn + 2H2nd • H,
A 0.2g sample of the alloy gave 0.23089 litre of H measured at 273 K and 1 atm pressure. What is the weight of Al in the alloy ?
(a) 0.18g
(b) 0.3 g
(C) 0.6 g
(d) 0.86g​

Answer 1

Answer:answer is option A

That is 0.18g

Explanation:

Answer 2

Answer:

The weight of Aluminium in the given alloy sample is equal to 0.18 grams.

Therefore, option (a) is correct.

Explanation:

Given , the chemical reaction of Aluminium and Zinc dissolved in HCl,

2Al  + 6HCl   →   AlCl₃  + 3H₂

Given, the weight of sample of alloy = 0.2g

Consider that 'w' is the weight of Al in the alloy sample. The weight of zinc in the sample (0.2 - w)g.

Given, the 2 moles of aluminium gives H₂ gas  = 3 moles

Then w/27 moles of aluminium will give $=\frac{3\times w}{2\times 27} =\frac{w}{18}\;moles$

The chemical reaction:

Zn + 2 HCl  →  ZnCl₂  +  H₂

1 moles of Zinc gives H₂ gas  = 1 moles

(0.2-w)/65 moles of Zinc will give  H₂ gas $=\frac{0.2- w}{65} \;moles$

The sample gives hydrogen gas = 0.23098 L

The number of moles of H₂ gas sample gave  $=\frac{0.23098}{22.4}=0.0103$

Therefore total number of moles  H₂ gas is equal to  sum of number of moles of H₂ gas  released by Al and Zinc.

$\frac{w}{18} +\frac{0.2-w}{65} = 0.0103$

$\frac{65w+(0.2)(18)-18w}{65\times 18}=0.0103$

$47w+ 3.6 = 12.064$

$47w =8.464$

$w=0.18g$

Therefore, the weight of Al in the alloy is equal to 0.18g.

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