найти массовую долю элементов (всех) в соединених?

P2O5

HNo3

AsH3

задачка

Мr(P2O5) = 2Аr(P) + 5Аr(O);Мr(P2O5) = 2 • 31 + 5 • 16 = 142;W(P) = 62/142 = 0, 44 = 44 %

W (O) = 80/142 = 0,56 = 56 %

Мr(HNO3) = Аr(H) + Аr(N) + 3Аr(O) ;Мr(HNO3) = 1 + 14 + 3 • 16 = 63;W(H) = 1/63 = 0, 016 = 1,6 %

W (N) = 14/63 = 0, 22 = 22 %

W (O) = 48/63 = 0, 76 = 76%

Мr(AsH3) = Аr(As) + 3Аr(H) ;Мr(AsH3) = 75 + 3 • 1 = 78;W(As) = 75/78 = 0, 96 = 96 %

W (H) = 3/78 = 0, 038 = 3,8 %

№1P2O5     P=31*2=62,  O=16*5=80

80+62=142

142----100%

80------x

 x=(80*100)/142=56,3%(O)

100-56,3...=43.66 т.е.44%(P)

 ответ:P=43,66%

            O=56,3%

№2 HNO3     H=1, N=14,O=16*3=48

 1+14+48=63

63----100%

1-------x

x=(100*1)/63=1,58% т.е. 1,6%(H)

63-1=62

100%-1,58...=98,41%

62------98,41%

14-------x

 x=(14*98,41)/62=22,22%(N)

98,41-22,22=76,19%(O)

ответ:H=1,6% или 1,58%

           N=22,22%

           O=76,19%

 №3 AsH3    As=75, H=1*3=3

75+3=78

78-----100%

75-----x

x=(75*100)/78=96,15%

100-96,15...=3,84%

ответ:As=96,15%

          H=3,84%