Hydrogen gas can be prepared by reaction of zinc metal with aqueous HCl : Zn(s)+2HCl(aq)â†’ZnCl2(aq)+H2(g) 1)How many liters...

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Hydrogen gas can be prepared by reaction of zinc metal withaqueous HCl :
Zn(s)+2HCl(aq)â†’ZnCl2(aq)+H2(g)
1)How many liters of H2 would be formed at 592 mm Hg and 17 âˆ˜Cif 25.5 g of zinc was allowed to react?
2)How many grams of zinc would you start with if you wanted toprepare 4.60 L of H2 at 294 mm Hg and 34.5 âˆ˜C ?

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4.2 Ratings (467 Votes)

Â Â Zn(s)+2HCl(aq) --- > ZnCl2(aq)+H2(g)

1 mol Zn = 1 mol H2

1) no of mol of Zn reacted = 25.5/65.4 = 0.4 mol

Â Â no of mol of H2 liberated = 0.4 mol

Â Â volume of H2 liberated = nRT/P

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.4*0.0821* 290.15 / (592/760)

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 12.23 L

2) no of mol of H2 leberated = PV/RT

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = (294/760)*4.6/(0.0821*307.65)

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.07 mol

Â Â Â Â no of mol of Zn must be reacted = 0.07 mol

Â Â Â Â mass of Zn reacted = n*M

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.07*65.4

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 4.6 g

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Hydrogen gas can be prepared by reaction of zinc metal with aqueous HCl : Zn(s)+2HCl(aq)â†’ZnCl2(aq)+H2(g) 1)How many liters...

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