Question

Problema 9 va roooog mai ales punctul b

Answer 1

$a)~\rho=\cfrac{m_s}{V} \Rightarrow m_s=\rho \cdot V\Rightarrow m_s=0,8\cdot 400\Rightarrow m_s=320~g\\\\ c=\cfrac{m_d}{m_s} \cdot 100 \Rightarrow 100m_d=c\cdot m_s\Rightarrow m_d=\cfrac{c\cdot m_s}{100} \Rightarrow m_d=\cfrac{96 \cdot 320}{100}\Rightarrow \\\\\Rightarrow m_d=\cfrac{30720}{100} \Rightarrow m_d= 307,2~g$

$M_C__2H_6O}=2\cdot A_C+6\cdot A_H+1\cdot A_O=2 \cdot 12+6\cdot 1+1\cdot 16=\\\\=24+6+16=46~g/mol\\\\n=\cfrac{307,2}{46} \Rightarrow n=6,678~moli \\\\ V=n \cdot V_m\Rightarrow V=6,678 \cdot 22,4 \Rightarrow V\simeq149,6~L$

$b)\boxed{1}~1 \cdot 22,4~L~C_2H_4.....................................3 \cdot 22,4~L~O_2\\~~~~~~~~149,6~L~C_2H_4........................................x~L~O_2\\\\x=\cfrac{3 \cdot 22,4\cdot 149,6}{22,4}\Rightarrow x=448,8~L~O_2\\\\V=\cfrac{100}{20} \cdot V_{O_2}\Rightarrow V=5 \cdot 448,8 \Rightarrow V=2244~L~aer$

$\boxed{2}~~22,4~L~C_2H_4...................................2~moli~CO_2\\~~~~~149,6~L~C_2H_4...................................x~moli~CO_2\\\\ x=\cfrac{2 \cdot 149,6}{22,4} \Rightarrow x=\cfrac{299,2}{22,4}\Rightarrow x= 13,35 ~moli~CO_2;~~~~~~y=13,35~moli~H_2O\\\\ n=13,35\cdot 2=26,7\\\\\%CO_2=\cfrac{100 \cdot 13,35}{26,7} =50\%;~~~~~\%H_2O=50\%$