Đáp án:

\({V_{{O_2}}} = 2,8{\text{ lit; }}{{\text{m}}_{{P_2}{O_5}}} = 7,1{\text{ gam; }}{{\text{m}}_{KMn{O_4}}} = 46,47{\text{ gam}}\)

Giải thích các bước giải:

Phản ứng xảy ra:

\(4P + 5{O_2}\xrightarrow{{}}2{P_2}{O_5}\)

Ta có: \({n_P} = \frac{{3,1}}{{31}} = 0,1{\text{ mol}} \to {{\text{n}}_{{O_2}}} = \frac{5}{4}{n_P} = 0,125{\text{ mol}} \to {{\text{V}}_{{O_2}}} = 0,125.22,4 = 2,8{\text{ lít}}\)

\({n_{{P_2}{O_5}}} = \frac{1}{2}{n_P} = 0,05{\text{ mol}} \to {{\text{m}}_{{P_2}{O_5}}} = 0,05.(31.2 + 16.5) = 7,1{\text{ gam}}\)

\(2KMn{O_4}\xrightarrow{{}}{K_2}Mn{O_4} + Mn{O_2} + {O_2}\)

Vì quá trình thu khí hao thụt 15%

\(\to {n_{{O_2}{\text{ tạo ra}}}} = \frac{{0,125}}{{85\% }} = \frac{5}{{34}}{\text{ mol}} \to {{\text{n}}_{KMn{O_4}}} = 2{n_{{O_2}}} = \frac{5}{{17}}{\text{ mol}} \to {{\text{m}}_{KMn{O_4}}} = \frac{5}{{17}}.(39 + 55 + 16.4) = 46,47{\text{ gam}}\)

$n_P=3,1/31=0,1mol$

$4P+5O_2\overset{t^o}\to 2P_2O_5$

a/Theo pt :

$n_{O_2}=5/4.0,1=0,125mol$

$⇒V_{O_2}=0,125.22,4=2,8l$

b/Theo pt :

$n_{P_2O_5}=1/2.n_P=1/2.0,1=0,05mol$

$⇒m_{P_2O_5}=0,05.142=7,1g$

$c/2KMnO_4\overset{t^o}\to K_2MnO_4+MnO_2+O_2$

Theo pt :

$n_{KMnO_4}=2.n_{O_2}=2.0,125=0,25mol$

Vì bị hao hụt 15% nên

$⇒n_{KMnO_4}=0,25.115\%=0,2875mol$

$⇒m_{KMnO_4}=0,2875.158=45,425g$