дано
m техн(Al) = 33.75 g
W(прим) = 20%
m(ppaH2SO4) = 784 g
W(H2SO4) = 15%
m(Al2(SO4)3)-?
m чист(Al) = 33.75 - (33.75*20% / 100%) = 27 g
M(Al) = 27 g/mol
n(Al) = m(Al) / M(AL) = 27 /27 = 1 mol
m(H2SO4) = 784 * 15% / 100% = 117.6 g
M(H2SO4) = 98 g/mol
n(H2SO4) = m(HsSO4) / M(H2SO4) = 117.6 / 98 = 1.2 mol
n(Al) < n(H2SO4
2Al+3H2SO4-->Al2(SO4)3+3H2↑
M(Al2(SO4)3) = 278 g/mol
2n(Al) = n(Al2(SO4)3) = 1 mol
n(AL2(SO4)3) = 1/2 = 0.5 mol
m(Al2(SO4)3) = n*M = 0.5*278 = 139 g
ответ 139 г
дано
m техн(Al) = 33.75 g
W(прим) = 20%
m(ppaH2SO4) = 784 g
W(H2SO4) = 15%
m(Al2(SO4)3)-?
m чист(Al) = 33.75 - (33.75*20% / 100%) = 27 g
M(Al) = 27 g/mol
n(Al) = m(Al) / M(AL) = 27 /27 = 1 mol
m(H2SO4) = 784 * 15% / 100% = 117.6 g
M(H2SO4) = 98 g/mol
n(H2SO4) = m(HsSO4) / M(H2SO4) = 117.6 / 98 = 1.2 mol
n(Al) < n(H2SO4
2Al+3H2SO4-->Al2(SO4)3+3H2↑
M(Al2(SO4)3) = 278 g/mol
2n(Al) = n(Al2(SO4)3) = 1 mol
n(AL2(SO4)3) = 1/2 = 0.5 mol
m(Al2(SO4)3) = n*M = 0.5*278 = 139 g
ответ 139 г