даноm(HCOH) = 20gm(Ag2O) = 23.2 g--------------------------m(Ag)-?M(HCOH) = 30 g/moln(HCOH) = m/M = 20 / 30 = 0.67 molM(Ag2O) = 232 g/moln(Ag2O) = m/M = 23.2 / 232 = 0.1 moln(HCOH) > n(Ag2O) M(Ag) = 108 g/moln(Ag2O) = n(Ag) = 0.1 molm(Ag) = n*M = 0.1 * 108 = 10.8 gответ 10.8 гHCOH+Ag2O-->HCOOH+2Ag