The amount of zinc required to produce 224 mL of H2 at STP on treatment with dilute H2 SO4 will be(Zn =65)

Question

The amount of zinc required to produce 224 mL of H2 at STP on treatment with dilute H2 SO4 will be(Zn =65) (A) 65 g (B) 0.065 g (C) 0.65 g (D) 6.5 g

Tardigrade Admin · Accepted Answer

Zn + H 2 SO 4(due) longrightarrow Zn SO 4+ H 2 I mole of zn react to give 1 mole of hydroge 220 4 ml of H 2 no of mole = (v at STP(ml)/22400) = (224/22400) =(1/100) mole no of moles = ( textmass/ textatomic mass) =(1/100)mole =( textmass/65 g/mole) mass = 0.65 g

Q. The amount of zinc required to produce 224mL of H2​ at STP on treatment with dilute H2​SO4​ will be(Zn =65)

65 g

0.065 g

0.65 g

6.5 g

Zn+H2​SO4​(due)⟶ZnSO4​+H2​ I mole of zn react to give 1 mole of hydroge 2204ml of H2​ no of mole =22400vatSTP(ml)​ =22400224​=1001​ mole no of moles =atomic massmass​ =1001​mole=65g/molemass​ mass =0.65g

Q. The amount of zinc required to produce $224 m L$ of $H_{2}$ at STP on treatment with dilute $H_{2} S O_{4}$ will be(Zn =65)