chx thỏa mãn với câu trả lời của mik???
a, \(m_C=1.95\%=0,95\left(kg\right)=950\left(g\right)\)
\(\rightarrow n_C=\frac{475}{6}\left(mol\right)\)
\(PTHH:C+O_2\underrightarrow{^{to}}CO_2\)
\(\rightarrow n_{O2}=n_{CO2}=\frac{475}{6}\left(mol\right)\)
\(\rightarrow V_{O2}=\frac{5320}{6}\left(l\right)\rightarrow V_{kk}=V_{O2}:20\%=8866,67\left(l\right)\)
b, \(V_{CO2}=1773,3\left(l\right)\)
mC = 95%*1= 0.95 kg = 950 g
nC = 475/6 mol
C + O2 -to-> CO2
475/6_475/6__475/6
Vkk = 5VO2 = 475/6*5*22.4 = 8866.67 (l)
VCO2 = 475/6*22.4 = 1773.33 l