Ответ:
дано
m техн(CaC2) = 1.2 g
m(C2H2Br4) = 5.19 g
---------------------
W(CaC2) - ?
CaC2+H2O-->Ca(OH)2+C2H2
C2H2+2Br2-->C2H2Br4
M(C2H2Br4) = 346 g/mol
n(C2H2Br4) = m/M = 5.19 / 346 = 0.015 mol
n(CaC2) = n(C2H2Br4) = 0.015 mol
M(CaC2) = 64 g/mol
m чист (CaC2) = n*M = 0.015 * 64 = 0.96 g
W(CaC2) = m чист (CaC2) / m техн(CaC2) *100% = 0.96 / 1.2 * 100% = 80%
ответ 80%
Объяснение:
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Answers & Comments
Ответ:
дано
m техн(CaC2) = 1.2 g
m(C2H2Br4) = 5.19 g
---------------------
W(CaC2) - ?
CaC2+H2O-->Ca(OH)2+C2H2
C2H2+2Br2-->C2H2Br4
M(C2H2Br4) = 346 g/mol
n(C2H2Br4) = m/M = 5.19 / 346 = 0.015 mol
n(CaC2) = n(C2H2Br4) = 0.015 mol
M(CaC2) = 64 g/mol
m чист (CaC2) = n*M = 0.015 * 64 = 0.96 g
W(CaC2) = m чист (CaC2) / m техн(CaC2) *100% = 0.96 / 1.2 * 100% = 80%
ответ 80%
Объяснение: