Объяснение:
Mr(FeO) = 56+16=72
W(Fe) = Ar(Fe) *n / Mr(FeO) *100% = 56 *1 / 72 *100% =78%
W(O) = Ar(O) *n / Mr(FeO) *100% = 16*1 / 72 *100% =22%
Mr(P2O5) = 31*2 + 16*5 = 142
W(P) = Ar(P)*n / Mr(P2O5) *100% = 31*2 / 142 *100%=44%
W(O) = Ar(O) *n / Mr(P2O5) *100% = 16*5 / 142 *100% = 56%
Mr(Na2CO3) = 23*2 + 12 + 16*3 = 106
W(Na) = Ar(Nа) *n / Mr(Na2CO3) *100% = 23*2 / 106 *100% = 43%
W(C) = Ar(C) *n / Mr(Na2CO3) *100% = 12*1 / 106 *100% = 11%
W(O) = Ar(O) *n / Mr(Na2CO3) *100% = 16*3 / 106 *100% =46%
Mr(Al2(SO4)3) = 27*2 + 32*3 + 16*12 = 342
W(Al) = Ar(Al) *n / Mr(Al2(SO4)3 *100% = 27*2 / 342 *100% = 16%
W(S) = Ar(S) *n / Mr(Al2(SO4)3 *100% = 32*3 / 342 *100% = 28%
W(O) = Ar(O) * n / Mr(Al2(SO4)3 *100% = 16*12 / 342 *100% = 56%
дано
m(CH3CL)=2 g
V(прак C2H6)=355.2 ml = 0.3552 L
η-?
2g X L
2CH3CL+2Na-->C2H6+2NaCL M(CH3CL)=50.5 g/mol , Vm=22.4 L/mol
50.5 22.4
X=2*22.4 / 50.5 = 0.887 L
η = V(практ) / V(теор)*100% = 0.3552 / 0.887 * 100% = 40 %
ответ 40%
2. дано
М(CH4)=11.2 L
η = 60%
m(CBr4)-?
CH4+4Br2-->CBr4+4HBr
n(CH4)=V / Vm = 11.2 / 22.4 = 0.5 mol
n(CH4)=n(CBr4)= 0.5 mol
M(CBr4)= 332 g / mol
m(CBr4)= n*M = 0.5*332 = 166 g - масса теорит
m(CBr4)=166 * 60% / 100% = 99.6 g - масса практич
ответ 99.6 г