дано
m техн (Al) = 6 g
W(прим) = 10%
m(S) = ?
m чист (AL) = 6 -(6*10% / 100%) = 5.4 g
2Al+3S-->Al2S3
M(Al) = 27 g/mol
n(Al) = m(Al) / M(Al) =5.4 / 27 = 0.2 mol
2n(Al) = 3n(S)
n(S) =2*0.2 / 3 = 0.13 mol
M(S) = 32 g/mol
m(S) = n*M = 0.13 * 32 = 4.16 g
ответ 4.16 гр
Объяснение:
дано
m техн (Al) = 6 g
W(прим) = 10%
m(S) = ?
m чист (AL) = 6 -(6*10% / 100%) = 5.4 g
2Al+3S-->Al2S3
M(Al) = 27 g/mol
n(Al) = m(Al) / M(Al) =5.4 / 27 = 0.2 mol
2n(Al) = 3n(S)
n(S) =2*0.2 / 3 = 0.13 mol
M(S) = 32 g/mol
m(S) = n*M = 0.13 * 32 = 4.16 g
ответ 4.16 гр
Объяснение: