14,4 г. карбида алюминия растворили в 140 г. 9,8% раствора серной кислоты. Определить...

Дано
m(Al4C3)= 14.4 g
m(ppa H2SO4) = 140 g
W(H2SO4)= 9.8 %
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m(CH4)-?
V(CH4)-?
m(H2SO4) = m(ppa H2SO4) * W(H2SO4) / 100%
m(ppa H2SO4) = 140*9.8% / 100% = 13.72 g
M(Al4C3) = 144 g/mol
n(Al4C3)= m/M = 14.4 / 144 = 0.1 mol
M(H2SO4) = 98g/mol
n(H2SO4)= m/M = 13.72 / 98 = 1.4 mol
n(Al3C4) < n(H2SO4)
14.4 X
Al4C3+6H2SO4-->3CH4 +2 Al2(SO4)3 M(CH4) = 16 g/mol
144 3*16
X= 14.4 * 3*16 / 144 = 4.8 g
V(CH4) = m(CH4) * Vm / M(CH4) = 4.8 * 22.4 / 16 = 6.72 L
ответ 4.8 г, 6.72 л