Ответ:

Объяснение:

M(Ca(HCO3)2) = 40 + (1 + 12 + 16 * 3) * 2 = 162 г/моль

w(Ca) = Ar(Са) * n/M(Ca(HCO3)2) * 100% = 40/162* 100% = 24,7%

w(Н) = Ar(Н) * n/M(Ca(HCO3)2) * 100% = 1 * 2/162 * 100% = 1,2%

w(C) = Ar(С) * n/M(Ca(HCO3)2) * 100% = 12 * 2/162 * 100% = 14,8%

w(О) = Ar(О) * n/M(Ca(HCO3)2) * 100% = 16 * 6/162 * 100% = 59,3%

М(Fe(OH)2NO3 = 56 +(16 + 1) * 2 + 14 + 16 * 3 = 152 г/моль

w(Fe) = Ar(Fe) * n/M(Fe(OH)2NO3) * 100% = 56/152* 100% = 36,84%

w(О) = Ar(О) * n/M(Fe(OH)2NO3) * 100% = 16 * 5/152 * 100% = 52,64%

w(Н) = Ar(Н) * n/M(Fe(OH)2NO3) * 100% = 1 * 2/152 * 100% = 1,31%

w(N) = Ar(N) * n/M(Fe(OH)2NO3) * 100% = 14/152 * 100% = 9,21%

М(KAl(SO4)2)= 39 + 27 + (32 + 16 * 4) * 2 = 258 г/моль

w(К) = Ar(К) * n/M(KAl(SO4)2) * 100% = 39/258 * 100% = 15,12%

w(Al) = Ar(Al) * n/M(KAl(SO4)2) * 100% = 27/258 * 100% = 10,46%

w(S) = Ar(S) * n/M(KAl(SO4)2) * 100% = 32 * 2/258 * 100% = 24,81%

w(О) = Ar(О) * n/M(KAl(SO4)2) * 100% = 128/258 * 100% = 49,61%