Ca(NO3)2+Na2CO3=NaNO3+CaCO3<span>↓ белый осадок</span>
Написаная _ означает индекс
Восстановительные свойства : 4NH_3 +3O_2=2N_2 + 6 H_2O
Реакция Присоединения : NH_3+H_2O-----> (NH_4)+OH
<span>H2SO4 + 2 LiOH = Li2SO4 + 2 H2O
</span>
Ответ:
Объяснение:
Mr(H3PO4) = 1*3 + 31+16*4 =98
W(H) = Ar(H) *n / Mr(H3PO4) *100% = 1*3 / 98 *100% =3%
W(P) = Ar(P) *n / Mr(H3PO4) *100% = 31*1 / 98 *100% =32%
W(O) = Ar(O) *n / Mr(H3PO4) *100% = 16*4 / 98 *100% =65%
2) Mr(SO3) = 32+16*3 = 80
W(S) = Ar(S) *n / Mr(SO3) *100% = 32*1 / 80 *100% =40%
W(O) = Ar(O) *n / Mr(SO3) *100% = 16*3 / 80 *100% = 60%
3) Mr(H2SO4) = 1*2 + 32+16*4 = 98
W(H) = Ar(H) *n / Mr(H2SO4) *100% = 1*2 / 98 *100% =2%
W(S) = Ar(S) *n / Mr(H2SO4) *100% = 32*1 / 98 *100% =33%
W(O) = Ar(O) *n / Mr(H2SO4) *100% = 16*4 / 98 *100% =65%
дано
m техн. ((NH4)2CO3) = 500 g
W(пр.) = 18%
m техн.(CaCL2) = 200 g
W( пр.) = 5%
-----------------------
m(CaCO3)-?
m чист (NH4)2CO3 = 500 - (500 * 18% / 100%) = 410 g
M((NH4)2CO3) = 96 g/mol
n((NH4)CO3) = m/M = 410 / 96 = 4.27 mol
m чист (CaCL2)= 200 - (200*5% / 100% ) = 190 g
M(CaCL2) = 111 g/mol
n(CaCL2) = m/M = 190 / 111 = 1.71 mol
n((NH4)2CO3) > n(CaCL2)
(NH4)2CO3+CaCL2 -->CaCO3+2NH4CL
n(CaCL2) = n(CaCO3) = 1.71 mol
M(CaCO3) = 100 g/mol
m(CaCO3) = n*M = 1.71 * 100 = 171 g
ответ 171 г