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zad1.
2H₂ + O₂ →2H₂O
2mole---->2mole
4g------>36g
200g---->x
x=1800g
odp
otrzymamy 1800g H2O
zad2
mAl=27u
mAl2S3=2*27u+3*32u=150u
2Al + 3S---->Al2S3
54g Al------>150g Al2S3
xg Al----->200g Al2S3
x=54*200/150u
x=72g Al
odp
potrzeba 72g Al
pozdr i liczę na naj:)