Question

what amount of magnesium is required to react with dilute H2 S o4 to produce 11.2 L of H2 at STP. ( Mg = 24)​

Answer 1

Answer:

12 g

Explanation:

Look,

given is , Mg + H2SO4 -->. MgSO4 + H2

this means 1 mole of mg reacts with 1 mole of hydrochloric acid

no.of mole in 11.2 L of H2 = 11.2/22.4

= 0.5 mol

now 0.5 H2 will reacts with 0.5 mole of mg

no. of mole mg = 0.5

mass/ atomic mass = 0.5

mass = 24× 0.5

= 12 g

Answer 2

The correct answer is 12 grams

GIVEN

Volume of H₂ produced at STP = 11.2L

TO FIND

Amount of Magnesium required.

SOLUTION

We can simply solve the above problem as follows,

The reaction between Mg and H₂SO₄ can be represented as,

Mg + H₂SO₄ ------------> MgSO₄ + H₂

We can observe that 1 mole of Mg gives 1 mole of H₂ gas.

Amount of H₂ gas produced = 11.2 L at STP

We know that,

1 mole of any gas at STP = 22.4 l

This means,

11.2 L of H₂ at stp = 0.5 moles

So, Moles of Mg reacts required for 0.5 moles of H₂ = 0.5 moles

We know that,

Moles = given weight / molecular weight

0.5 = given weight/24

Given weight = 0.5 × 24 = 12 grams

Hence, The correct answer is 12 grams

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