дано
m(ppa Na2S) = 50 g
W(Na2S) = 5%
+HCL
m(Na2S) = 50 * 5% / 100% = 2.5 g
Na2S+2HCL-->2NaCL+H2S↑
M(Na2S) =78 g/mol
n(Na2S) = n/M = 2.5 / 78 = 0.32 mol
n(Na2S) = n(H2S) = 0.32 mol
V(H2S) = n*Vm =0.32 *22.4 = 7.168 L
ответ 7.168 л
Объяснение:
дано
m(ppa Na2S) = 50 g
W(Na2S) = 5%
+HCL
m(Na2S) = 50 * 5% / 100% = 2.5 g
Na2S+2HCL-->2NaCL+H2S↑
M(Na2S) =78 g/mol
n(Na2S) = n/M = 2.5 / 78 = 0.32 mol
n(Na2S) = n(H2S) = 0.32 mol
V(H2S) = n*Vm =0.32 *22.4 = 7.168 L
ответ 7.168 л
Объяснение: