<span>б. С+2Н2=СН4
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Дано
m = 780 г
w(примесей) = 25% = 0.25
Решение
w(NaCl) = 1 - w(примесей) = 1 - 0.25 = 0.75
m(NaCl) = mw(NaCl) = 780 * 0.75 = 585 г
n(NaCl) = m(NaCl) / M(NaCl) = 585 / (23 + 35.5) = 585 / 58.5 = 10 моль
H2SO4 + 2NaCl = Na2SO4 + 2HCl
n(HCl) = n(NaCl) = 10 моль
V(HCl) = n(HCl) * 22.4 = 10 * 22.4 = 224 дм3
Ответ: 224 дм3
<span>Ba2 + 4H2O =2 Ba(OH)2 + 2H2↑</span>
1СaО+ 2HNO3->Ca(NO3)2+H2O
2..ZnCL2+2NaOH->2NaCL+2Zn(OH)2
3 CL2+KI->L2+KCI
Элемент L??
4.2AL+CuBr2->2ALBr+Cu
52AL(OH)3->AL2O3+3H2O
6Cu(OH)2->CuO+H2
7 4AL+3C->AL4C3
Сl2 + H2O -> HCl + HClO
2Al +3 I2 -> 2AlI3
MgO + 2HBr -> MgBr2 + H2O
2KBr + Cl2 -> 2KCl + Br2
AgNO3 + HCl -> AgCl + HNO3