Zapremina gasa

Submitted by dejanbr1 on Tue, 10/29/2013 - 20:06
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1.Kolika se zapremina gasa oslobodi potpunom reakcijom dovoljne kolicine Al sa 150cm3 30% H2SO4,gustine 1,16g/cm3?

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milicaradenkovic - Tue, 10/29/2013 - 22:01

2Al+3H2SO4--->Al2(SO4)3+3H2

V(H2SO4)=150 cm3

w(H2SO4)=30 % a ro=1.16 g/cm3

mr=ro*V(H2SO4)=1.16*150=174 g

w(H2SO4)=m(H2SO4)/mr

m(H2SO4)=w(H2SO4)*mr=0.3*174=52.2 g

n(kiseline)=m(H2SO4)/Mr=52.2/98=0.533 mol

n(kiseline):n(H2)=3:3

n(H2)=3*n(kiseline)/3=3*0.533/3=0.533 mol

1 mol------------------------22.4 dm3

0.533 mol-------------------x dm3

x=0.533*22.4/1=11.94 dm3 H2