<u>1)2</u><em>Fe</em> + <u>6</u><em>H</em>2<em>S</em><em>O</em>4(конц.) = <em>Fe</em>2(<em>S</em><em>O</em>4)3 + <u>3</u><em>S</em><em>O</em>2↑ + <u>6</u><em>H</em>2<em />
1 б
2 б
3 б
4 в
5 б
Решения надеюсь не нужны. Долго писать
FeCl3+3NaOH=3NaCl + Fe(OH)3(осадок)
дано
m(Al) = 8.1 g
m(S) = 18 g
-------------------
m(Al2S3)-?
M(AL) = 27 g/mol
n(Al) = m/M = 8.1 / 27 = 0.3 mol
M(S) = 32 g/mol
n(S) = m/M = 18 / 32 = 0.56 mol
n(AL) < n(S)
2Al+3S -->Al2S3
2n(Al) = n(Al2S3)
n(AL2S3) = 0.3 / 2 = 0.15 mol
M(Al2S3) = 150 g/mol
m(Al2S3) = n*M = 0.15 * 150 = 22.5 g
ответ 22.5 г