6)
дано
m(ppa CuSO4) = 35 g
W(CuSO4) = 20%
m( ppa NaOH) = 20 g
W(NaOH) = 25 %
-----------------------------
m(Cu(OH)2)-?
m(CuSO4) = m(ppa CuSO4) * W(CuSO4) / 100% = 35*20% / 100% = 7 g
M(CuSO4) =160 g/mol
n(CuSO4) = n/M = 7 / 160 = 0.04375 mol
m(NaOH) = m(ppa NaOH) * W(NaOH) / 100% = 20*25% / 100% = 5 g
M(NaOH) = 40 g/mol
n(NaOH) = n/M = 5/ 40 = 0.125 mol
n(CuSO4) < n(NaOH)
CuSO4+2NaOH-->Cu(OH)2+Na2SO4
n(CuSO4) = Cu(OH)2 = 0.04375 mol
M(Cu(OH)2) = 98 g/mol
m(Cu(OH)2) = n*M = 0.04375 * 98 = 4.2875 g
ответ 4.2875
7)
дано
m техн(WO3) = 240 g
W(прим) = 5%
-----------------------
V(H2)-?
m чист(WO3) = m техн(WO3) - (m техн (WO3) * W(прим) ) / 100%
m чист (WO3)= 240 - (240*5%) / 100% = 228 g
2WO3+3H2-->2W+3H2O
M(WO3) = 232 g/mol
n(WO3) = m/M = 228 / 232 = 0.98 mol
2n(WO3) = 3n(H2)
n(H2) = 3*0.98 / 2 = 1.47 mol
V(H2) = n*Vm = 1.47 * 22.4 = 32.928 L
ответ 32.928 л