Question

tinh the tich khi oxi va the tich khong khi [dktc] can thiet de dot chay:1 mol cacbon;1.5 mol photpho.cho biet oxi chiem 20% the tich khong khi.

Answer 1

a) ta có PTHH: O₂ + C \(\underrightarrow{t^0}\) CO₂

theo gt: n_{C =1(mol)}

theo PTHH: n_O2=n_C=1(mol)

\(\Rightarrow\) V_{O2=1*22,4=22,4(lit)}

\(\Rightarrow\)V_kk=\(\frac{V_{O2}\cdot100}{20}=\frac{22,4\cdot100}{20}=112\left(lit\right)\)

b)ta có PTHH : 4P+5O₂\(\underrightarrow{t^0}\) 2P₂O₅

_{theo gt:} n_P=1,5(mol)

_{\(\Rightarrow\)}n_O2=5/4*n_P=5/4*1,5=1,875(mol)

\(\Rightarrow\)V_O2=22,4*1,875=42(lit)

\(\Rightarrow\)V_kk=\(\frac{42\cdot100}{20}=210\left(lit\right)\)