a)
Ecuația reacției chimice:
x 0.2 moli
CaC₂ + 2H₂O --(t°C)-> Ca(OH)₂ + C₂H₂
1 mol 1 mol
Rezolvare:
[tex]V_{C_2H_2}=3.36\ dm^3\\V_M=22.4\ dm^3\ (volum\ molar)\\n=\frac{V}{V_M}=\frac{3.36}{22.4}=0.15\ moli\ C_2H_2\ (Cp)\\\\25\%\ pierderi\implies{\eta}=75\%\ (randamentul\ reactiei)\\{\eta}=\frac{Cp}{Ct}*100\impliesCt=\frac{Cp*100}{{\eta}}=\frac{0.15*100}{75}=0.2\ moli\ C_2H_2\\\\n_{CaC_2}=x=0.2\ moli\\M_{CaC_2}=A_{Ca}+2*A_C=40+(2*12)=64\ g/mol\\\\n=\frac{m}M\implies\ m=n*M=0.2*64=12.8\ g\ CaC_2\ (m_{imp})\\P=80\%\ (puritatea\ carbidului)\\P=\frac{m_p}{m_{imp}}*100\implies\ m_p=\frac{80*12.8}{100}=10.24\ g[/tex]
b)
Ecuația reacției chimice:
0.15 moli x
HC ≡CH + 2Br₂ --(CCl₄)--> (Br₂)C-C(Br₂)
1 mol 2 moli
Rezolvare:
[tex]V_{C_2H_2}=3.36\ dm^3\\V_M=22.4\ dm^3\\n=\frac{V}{V_M}=\frac{3.36}{22.4}=0.15\ moli\ C_2H_2\\\\n_{Br_2}=x=0.3\ moli\\M_{Br_2}=160\ g/mol\\n=\frac{m}M\implies\ m=n*M=0.3*160=48\ g\ Br_2\ (m_d)\\\\C=4\%\ (conc.\ sol.\ de\ Br_2)\\C=\frac{m_d}{m_s}*100\implies\ m_s=\frac{m_d*100}{C}=\frac{48*100}{4}=1200\ g\ sol.\ Br_2[/tex]
c)
Ecuația reacției chimice:
x 0.375 moli
C₂H₂ + 5/2 O₂ ---(t°C)--> 2CO₂ + H₂O + Q
1 mol 2.5 moli
Rezolvare:
[tex]V_{aer}=42\ L=42\ dm^3\\V_{aer}=5*V_{O_2}\ (20\%)\implies\ V_{O_2}=\frac{V_{aer}}{5}=\frac{42}{5}=8.4\ L\\V_M=22.4\ dm^3\ (vol.\ molar)\\\\n=\frac{V}{V_M}=\frac{8.4}{22.4}=0.375\ moli\ O_2\\n_{C_2H_2}=x=0.15\ moli\\n=\frac{V}{V_M}\impliesV=n*V_M=0.15*22.4=3.36\ L\ C_2H_2[/tex]
