Hydrogen gas can be prepared by reaction of zinc metal with

Hydrogen gas can be prepared by reaction of zinc metal with aqueous HCl: Zn(s)+2HCl(aq)=ZnCl2(aq)+H2(g)
How many grams of zinc would you start with if you wanted to prepare 4.90L of H2 at 290 mmHg and 33.0 C?

Solution

pv=nrt

p=290/760=0.38atm

0.38*4.9=n*0.0821*306

n=0.074

so 0.074 moles of zn=0.074*65.38=4.84 gms

Hydrogen gas can be prepared by reaction of zinc metal with aqueous HCl: Zn(s)+2HCl(aq)=ZnCl2(aq)+H2(g) How many grams of zinc would you start with if you wante

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