V = 10 (л)
n=V/Vm, Vm - объем газа при н.у. = 22,4 (л/моль)
n=10/22.4 =0.45
М(NH3) = 14+3*1=17 (г/моль)
m= n*M = 0.45*17 = 7.65 (г)
CuCl2+2NaOH=Cu(OH)2↓+2NaCl
n(NaOH)=8/40=0.2 моль
n(Cu(OH)2)=0.5 n(NaOH)=0.1 моль
m(Cu(OH)2)=0.1*98=9.8 г.
3Zn + 4H₂SO₄<span>(конц)→<span> 3ZnSO</span></span>₄<span><span> + S + 4H</span></span>₂<span><span>O
</span></span>Cu + 2H₂SO₄<span>(конц.)→<span> CuSO</span></span>₄<span><span> + SO</span></span>₂ <span><span>+ 2H</span></span>₂<span><span>O
8Na + 5H</span></span>₂SO₄(конц.)= 4Na₂SO₄ + H₂S + 4H₂O
С + О2 = СО2
М(чистого С) =(М*V)/Vm= (12*336)/22,4 = 180 г.
w(массовая доля)=(m(C)/m(угля))*100%= (180/187,52)*100 = 95,989 = 96%
Дано
m(C3H7OH) = 60 kg = 60000 g
η(C3H6) -?
-----------------------------------
V практ(C3H6)-?
C3H7OH-->C3H6+H2O
M(C3H7OH) = 60 g/mol
n(C3H7OH) = m/M = 60000 / 60 = 10000 mol
n(C3H7OH) = n(C3H6) = 10000mol
V теор (C3H6) = n*Vm = 10000 * 22.4 = 224000 L
V практ(C3H6) = V(C3H6) * η / 100\% = 224000 * 70\% / 100\% = 156800 L
ответ 156800 л