a)
Phản ứng xảy ra:
\(C + {O_2}\xrightarrow{{{t^o}}}C{O_2}\)
Ta có:
\({n_C} = {n_{{O_2}}} = \frac{3}{{12}} = 0,25{\text{ mol}} \to {{\text{m}}_{{O_2}}} = 0,25.32 = 8{\text{ gam;}}{{\text{V}}_{{O_2}}} = 0,25.22,4 = 5,6{\text{ lít}} \to {{\text{V}}_{kk}} = 5{V_{{O_2}}} = 28{\text{ lít}}\)
b)
Phản ứng xảy ra:
\({C_4}{H_{10}} + \frac{{13}}{2}{O_2}\xrightarrow{{{t^o}}}4C{O_2} + 5{H_2}O\)
Ta có:
\({n_{{C_4}{H_{10}}}} = \frac{{112}}{{22,4}} = 5{\text{ mol}} \to {{\text{n}}_{C{O_2}}} = \frac{{13}}{2}{n_{{C_4}{H_{10}}}} = 32,5{\text{ mol}} \to {{\text{m}}_{{O_2}}} = 32,5.32 = 1040{\text{ gam}}\)
\({V_{{O_2}}} = 32,5.22,4 = 728{\text{ lít}} \to {{\text{V}}_{kk}} = 728.5 = 3640{\text{ lít}}\)
c)Phản ứng xảy ra:
\(C + {O_2}\xrightarrow{{{t^o}}}C{O_2}\)
Ta có:
\({m_C} = 1000.95\% = 950{\text{ kg}} \to {{\text{n}}_C} = \frac{{950}}{{12}} = {n_{{O_2}}} \to {m_{{O_2}}} = \frac{{950}}{{12}}.32 = 2533,3kg\)
\( \to {V_{{O_2}}} = \frac{{950}}{{12}}.22,4 = \frac{{5320}}{3}{m^3} \to {V_{kk}} = 5{V_{{O_2}}} = 8866,67{m^3}\)
