дано
m(Mg) = 7.2 g
V(CL2) = 7.8 L
-----------------
m(MgCL2)=?
M(Mg) = 24 g/mol
n(Mg) = m/M = 7.2 / 24 = 0.3 mol
n(CL2) = V(CL2) / Vm = 7.8 / 22.4 = 0.35 mol
n(Mg) < n(CL2)
Mg+CL2-->MgCL2
n(Mg) = n(MgCL2) = 0.3 mol
M(MgCL2) = 95 g/mol
m(MgCL2) = n*M =0.3*95 = 28.5 g
ответ 28.5 г
2)
дано
m(ppa Na2CO3) = 120 g
W(Na2CO3) = 40 %
m(HCL) = 35 g
------------------------
V(CO2)-?
m(Na2CO3) = 120 * 40% / 100% = 48 g
M(Na2CO3) = 106 g/mol
n(Na2CO3) = m/M = 48 / 106 = 0.45 mol
M(HCL) = 36.5 g/mol
n(HCL) = m/M = 35 / 36.5 = 0.99 mol
n(Na2CO3) < n(HCL)
Na2CO3+2HCL-->2NaCL+H2O+CO2
n(Na2CO3) = n(CO2) = 0.45 mol
V(CO2) = n*Vm = 0.45 * 22.4 = 10.08 L
ответ 10.08 л
Mg+2HCl=MgCl2+H2
Mg(0)-2e=Mg(+2). 1. ок-ие
в-ль
2H(+)+2e=H2(0). 1. в-ие
ок-ль
Fe+CuSO4=FeSO4+Cu
Fe(0)-2e=Fe(+2). 1. ок-ие
в-ль
Cu(+2)+2e=Cu(0). 1. в-ие
ок-ль
2Al+Fe2O3=Al2O3+2Fe
Al(0)-3e=Al(+3). 1. ок-ие
в-ль
Fe(+3)+3e=Fe(0). 1. в-ие
ок-ль
...................................