2Al+3S=Al2S3
29,7--X
54--96 X=52,8 (значит сера в избытке)
Решаем задачу по недостатку
29,7--Y
54--150 Y=82,5 (масса соли)
Ответ:82,5
180/32=5,625моль
5.625×64=360г
<span>K</span>₂<span>O + 2HCl = 2KCl + H</span>₂<span>O
NaOH + HCl = NaCl + H</span>₂<span>O
Al</span>₂<span>O</span>₃<span> + 6HCl = 2AlCl</span>₃<span> + 3H</span>₂<span>O
Zn(OH)</span>₂<span> + 2HCl = ZnCl</span>₂<span> + 2H</span>₂<span>O
ZnO + 2HCl = ZnCl</span>₂<span> + H</span>₂<span>O
CaO + 2HCl = CaCl</span>₂<span> + H</span>₂<span>O
2Na + 2HCl = 2NaCl + H</span>₂
N=n*Vm
n(N2) = 2*22,4= 44.8л
n(O2)= 3*22.4=67,2 л
44,8+67,2=112л