- Số mol \(P_2O_5:n_{P_2O_5}=\dfrac{m}{M}=\dfrac{71}{142}=0,5\left(mol\right)\)
\(-PTHH:4P+5O_2\underrightarrow{t^o}2P_2O_5\)
\(\left(mol\right)\) \(4\) \(5\) \(2\)
\(\left(mol\right)\) \(1\) \(1,25\) \(0,5\)
a. Thể tích Oxi cần dùng là:
\(V_{O_2}=n.22,4=1,25.22,4=28\left(l\right)\)
b. Khối lượng Photpho đã dùng là:
\(m_P=n.M=1.31=31\left(g\right)\)
PTHH: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\\ 1mol:1,25mol\leftarrow0,5mol\)
\(n_{P_2O_5}=\dfrac{71}{142}=0,5\left(mol\right)\)
a. \(V_{O_2}=1,25.22,4=28\left(l\right)\)
b. \(m_P=1.31=31\left(g\right)\)
PTHH:4P+5O2\(\rightarrow\)2P2O5
a)Theo đề, ta có:nP2O5=\(\dfrac{71}{142}\) =0,5(mol)
Theo PTHH:\(\dfrac{n_{O_2}}{n_{P_2O_5}}\)=\(\dfrac{5}{2}\)\(\Rightarrow\)2.nO2=5.nP2O5\(\Rightarrow\)nO2=\(\dfrac{5}{2}\).0,5=1,25(mol)
\(\Rightarrow\)vO2=1,25.22,4=28(l)
Vậy: VO2=28 l
b)Theo PTHH:\(\dfrac{n_P}{n_{P2O5}}\)=\(\dfrac{2}{1}\)\(\Rightarrow\)nP=2.nP205\(\Rightarrow\)nP=2.0,5=1(mol)
\(\Rightarrow\)mP=1.31=31(g)
Vậy:mP=31g
a) \(n_{P_2O_5}=\dfrac{71}{142}=0,5\left(mol\right)\)
PTHH: \(4P+5O_2\underrightarrow{t^0}2P_2O_5\)
Theo PTHH: \(n_{P_2O_5}:n_{O_2}=2:5\)
\(\Rightarrow n_{O_2}=n_{P_2O_5}.\dfrac{5}{2}=0,5.\dfrac{5}{2}=1,25\left(mol\right)\)
\(\Rightarrow V_{O_2}=1,25.22,4=28\left(l\right)\)
b) Theo PTHH: \(n_{P_2O_5}:n_P=2:4\)
\(\Rightarrow n_P=n_{P_2O_5}.2=0,5.2=1\left(mol\right)\)
\(\Rightarrow m_P=1.31=31\left(g\right)\)
