Ответ:

дано

m(CxHyOz) = 7.5 g

m(H2O) = 4.5 g

m(CO2) = 11 g

D(H2) = 15

----------------------

CxHyOz-?

M(CXHyOz) = D(H2) * 15 = 2* 15 = 30 g/moL

n(H) = 2*n(H2O) = 2 * 4.5 / 18 = 0.5 mol

n(C) = n(CO2) = m/M = 4 / 11 = 0.25 mol

m(H) = n(H) * M(H)  = 0.5 * 1 = 0.5 g

m(C) = n(C)*M(C)  = 0.25 * 12 = 3 g

m(O) = m(CxHyOz)  - m(C) + m(H) ) = 7.5 - (3 + 0.5 ) = 4 g

n(O) = m(O) / M(O) = 4 / 16 = 0.25 mol

n(C) : n(H) : n(O) = 0.25 : 0.5 : 0.25 = 1 : 2 : 1

HCOH

ответ 30 г/моль, метаналь

2.  дано

n(CH3OH) = 2 mol

+m(H2O) = 200 g

----------------

W(CH2O)-?

2CH3OH+O2-->2HCOH+2H2O

2n(CH3OH) = 2n(HCOH) = 2 mol

n(HCOH) =2 mol

M(HCOH) = 30 g/mol

m(HCOH) = n*M = 2* 30 = 60 g

находим массу образовавшийся воды

2n(CH3OH) = 2n(H2O)

n(H2O) = 2 mol

M(H2O) = 18 g/mol

m(H2O) = n*M = 2 * 18  = 36 g

находим общую массу воды

m( общ H2O) = 200 + 36  = 236 g

m(ppa HCOH) = m(HCOH) + m(общ H2O) = 60 + 236 = 296 g

W(HCOH) = m(HCOH) / m(ppa HCOH)*100% = 60 / 296 * 100% = 20.27%

ответ 20.27%

Объяснение: