Đáp án:

b) \({m_C} = 3,6{\text{ gam; }}{{\text{m}}_S} = 12,8{\text{ gam}}\)

c) \({{\text{V}}_{C{O_2}}} = 6,72{\text{ lít}}\)

d) \({V_{kk}} = 78,4{\text{ lít}}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(C + {O_2}\xrightarrow{{}}C{O_2}\)

\(S + {O_2}\xrightarrow{{}}S{O_2}\)

Ta có: \({n_{{O_2}}} = \frac{{15,68}}{{22,4}} = 0,7{\text{ mol; }}{{\text{n}}_{S{O_2}}} = \frac{{25,6}}{{32 + 16.2}} = 0,4{\text{ mol}}\)

Theo phản ứng: 

\({n_S} = {n_{S{O_2}}} = 0,4{\text{ mol; }}{{\text{n}}_{{O_2}}} = {n_C} + {n_S} \to {n_C} = 0,7 - 0,4 = 0,3{\text{ mol}}\)

\( \to {m_C} = 0,3.12 = 3,6{\text{ gam; }}{{\text{m}}_S} = 0,4.32 = 12,8{\text{ gam}}\)

Ta có: \({n_{C{O_2}}} = {n_C} = 0,3{\text{ mol}} \to {{\text{V}}_{C{O_2}}} = 0,3.22,4 = 6,72{\text{ lít}}\)

\({V_{kk}} = 5{V_{{O_2}}} = 5.15,68 = 78,4{\text{ lít}}\)

$n_{SO_2}=25,6/64=0,4mol$

$n_{O_2}=15,68/22,4=0,7mol$

$a/$

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

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

$b/$

Theo pt(2):

$n_S=n_{SO_2}=0,4mol$

$⇒n_{C}=0,7-0,4=0,3mol$

$⇒m_C=0,3.12=3,6g$

$m_S=0,4.32=12,8g$

$c/$

Theo pt (1):

$n_{CO_2}=n_C=0,3mol$

$⇒V_{CO_2}=0,3.22,4=6,72l$

$d/$

$V_{Kk}=5.22,4.0,7=78,4l$