Koliko ce se osloboditi mililitara gasa (normalni uslovi) rastvaranjem 540
mg aluminijuma u natrijum-hidroksidu?
2Al+2NaOH+6H2O--------------->2Na{Al(OH)4}+3H2
m(Al)=540 mg
n(Al)=m(Al)/Mr=540*10(-3)/27=0.02 mol
n(Al):n(H2)=2:3
n(H2)=n(Al)*3/2=0.02*3/2=0.03 mol
1 mol------------------22.4 dm3
0.03 mol--------------x dm3
x=0.03*22.4/1=0.672 dm3=672 cm3 H2
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Al+NaOH
2Al+2NaOH+6H2O--------------->2Na{Al(OH)4}+3H2
m(Al)=540 mg
n(Al)=m(Al)/Mr=540*10(-3)/27=0.02 mol
n(Al):n(H2)=2:3
n(H2)=n(Al)*3/2=0.02*3/2=0.03 mol
1 mol------------------22.4 dm3
0.03 mol--------------x dm3
x=0.03*22.4/1=0.672 dm3=672 cm3 H2