C-->CO2
хг 44,8л
C-->CO₂
12г 22,4л
m(C)=44,8л*12г\22,4л=24г
хг 45г
H2-->H2O
2г 18г
m(H)=45г*2г\18г=5г
СxHy
12x:y=24:5
x:y=2:5
C₂H₅ - простейшая формула
Значит алкан имеет формулу C₄H₁₀
2C₄H₁₀ + 13O₂ --> 8CO₂ + 10H₂O
n(CO₂)=44,8л\22,4 л\моль = 2 моль
n(C₄H₁₀)=1\4 n(CO₂)=0,5 моль
V(C₄H₁₀)=22,4 л\моль*0,5 моль=11,2л
<em>Ответ: C₄H₁₀ - бутан ; 11,2л (0,5 моль )</em>
120г Хл
Мg + 2CH3COOH = (CH3COO)2Mg + H2
120г 22,4л
х=120*22,4/120=22,4л
<span>C → CO</span>₂<span> → Na</span>₂<span>CO</span>₃<span> → CO</span>₂<span> → CaCO</span>₃
<span><span>
1) </span>C + O</span>₂<span> → CO</span>₂<span>;</span>
<span><span>2) </span>CO</span>₂<span> +
2NaOH →<span>
Na</span></span>₂<span><span>CO</span></span>₂<span><span> + H</span></span>₂<span><span>O;</span></span>
<span><span>3) </span><span>Na</span></span>₂<span><span>CO</span></span>₃<span><span> + 2HCl </span>→ 2NaCl + CO</span>₂<span>↑
+ H</span>₂<span>O;</span>
<span><span>4) </span><span>CO</span></span>₂<span><span> +
Ca(OH)</span></span>₂<span><span> </span>→ CaCO</span>₃<span>↓<span> + H</span></span>₂<span><span>O</span></span>