Đáp án:

 1. a = 14

V = 11,2l

V' = 2,8l

2. 

m = 3,6g

Giải thích các bước giải:

 1. \[m_{C}=10\cdot 60\%=6\ (gam)\Rightarrow n_{C}=\dfrac{6}{12}=0,5\ (mol)\\ m_{S}=10-6=4\ (gam)\Rightarrow n_{S}=\dfrac{4}{32}=0,125\ (mol) \]

\(C+O_2\xrightarrow{t^{\circ}} CO_2\\ S+O_2\xrightarrow{t{^\circ}}SO_2\\\to n_{O_2}=n_{C}+n_{S}=0,125+0,5=0,625\ (mol)\to a=0,625\cdot 22,4=14\ (l)\\ n_{CO_2}=n_{C}=0,5\to V_{CO_2}=0,5\cdot 22,4=11,2\ (l)\\ n_{SO_2}=n_{S}=0,125\ (mol)\to V_{SO_2}=0,125\cdot 22,4=2,8\ (l)\)

2. \[d_{M/H_2}=4,75 \Rightarrow\overline{M_X}=4,75\cdot 2=9,5\ (g/mol)\Rightarrow \dfrac{32n_{O_2}+2n_{H_2}}{n_{O_2}+n_{H_2}}=9,5\\ n_{O_2}+n_{H_2}=\dfrac{8,96}{22,4}=0,4\ (mol) \Rightarrow \begin{cases}n_{O_2}=0,1\\ n_{H_2}=0,3\end{cases}\]

\(2H_2+O_2\xrightarrow{t{^\circ}}2H_2O\\ \dfrac{0,3}{2}>\dfrac{0,1}{1}\)

\(\Rightarrow\)Sau phản ứng \(O_2\) hết

\(\Rightarrow n_{H_2O}=0,1\cdot 2=0,2\ (mol)\\ \Rightarrow m_{H_2O}=0,2\cdot 18=3,6\ (gam)\)

1.

$m_C=60\%.10=6g$

$⇒n_C=6/12=0,5mol$

$⇒m_S=10-6=4g⇒n_S=4/32=0,125mol$

$PTHH :$

$C+O_2\overset{t^o}\to CO_2(1)$

$S+O_2\overset{t^o}\to SO_2(2)$

Theo pt (1) và (2) :

$n_{O_2}=n_{O_2(1)}+n_{O_2(2)}=0,5+0,125=0,625mol$

$⇒a=V_{O_2}=0,625.22,4=14l$

$n_{CO_2}=n_C=0,5mol⇒V_{CO_2}=0,5.22,4=11,2l$

$n_{SO_2}=n_S=0,125mol⇒V_{SO_2}=0,125.22,4=2,8l$