Question

Mg +H2SO4. =>Mgso4 + h2lf 48g Mg react with 98 g H2so4 then amount of H2 produced is​

Answer 1

Answer:

No. of moles :

$\implies n_{Mg} = \dfrac{48}{24} = 2$

$\implies n_{H_2SO_4} = \dfrac{98}{2 \times 1+ 32 + 16 \times 4} = \dfrac{98}{98} = 1$

Limiting Reagent:

$\implies Mg = \dfrac{2}{1} = 2$

$\implies H_2SO_4 = \dfrac{1}{1} = 1$

Therefore,L.R is H₂SO₄.

By unitary method:

1 mol of H₂SO₄ produce moles of H₂ = 1 mol

1 mol of H₂ has = 2 g

Therefore, the amount of H₂ is produced is 2 g.