Question

Обчисліть масову частку Гідрогену у сполуках: H2S, HBr, PH3​

Answer 1

w(элемента)=n×Ar(элемента)/M(вещества)

w1(H) = 2×1/2×1+32 = 2/34 = 0,0588...(5,88%)

w2(H) = 1×1/1+80 = 1/81 = 0,0123..(1,23%)

w3(H) = 3×1/31+3×1= 3/34 = 0,0882...(8,82%)

Answer 2

Mr(H2S) = Ar(H) * N(H) + Ar(S) * N(S) = 1 * 2 + 32 * 1 = 34;

ω(H в H2S) = N(H в H2S) * Ar(H) * 100% / Mr(H2S) = 2 * 1 * 100% / 34 = 5,88%;

Mr(HBr) = Ar(H) * N(H) + Ar(Br) * N(Br) = 1 * 1 + 80 * 1 = 81;

ω(H в HBr) = N(H в HBr) * Ar(H) * 100% / Mr(HBr) = 1 * 1 * 100% / 81 = 1,23%;

Mr(PH3) = Ar(P) * N(P) + Ar(H) * N(H) = 31 * 1 + 1 * 3 = 34;

ω(H в PH3) = N(H в PH3) * Ar(H) * 100% / Mr(PH3) = 3 * 1 * 100% / 34 = 8,82%.