Question

Dot chay 12,4 gam photpho trong binh co 12,8 gam oxi

a)Viet PT phan ung xay ra

b)Hay cho biet chat nao con du? Du bao nhieu gam?

c) Tinh khoi luong san pham thu duoc sau phan ung

d) De co luong oxi dung cho phan ung tren thi phai nung bao nhieu gam thuoc tim?

Answer 1

\(n_P=\dfrac{12,4}{31}=0,4\left(mol\right)\)

\(n_{O_2}=\dfrac{12,8}{32}=0,4\left(mol\right)\)

a) 4P + 5O₂ \(\underrightarrow{to}\) 2P₂O₅ (1)

b) Theo PT1: \(n_P=\dfrac{4}{5}n_{O_2}\)

theo bài: \(n_P=n_{O_2}\)

Vì \(1>\dfrac{4}{5}\) ⇒ P dư

Theo PT1: \(n_Ppư=\dfrac{4}{5}n_{O_2}=\dfrac{4}{5}\times0,4=0,32\left(mol\right)\)

\(\Rightarrow n_Pdư=0,4-0,32=0,08\left(mol\right)\)

\(\Rightarrow m_Pdư=0,08\times31=2,48\left(g\right)\)

c) Theo PT1: \(n_{P_2O_5}=\dfrac{2}{5}n_{O_2}=\dfrac{2}{5}\times0,4=0,16\left(mol\right)\)

\(\Rightarrow m_{P_2O_5}=0,16\times142=22,72\left(g\right)\)

\(\Rightarrow m_{sp}=m_Pdư+m_{P_2O_5}=2,48+22,72=25,2\left(g\right)\)

d) 2KMnO₄ \(\underrightarrow{to}\) K₂MnO₄ + MnO₂ + O₂ (2)

Theo PT2: \(n_{KMnO_4}=2n_{O_2}=2\times0,4=0,8\left(mol\right)\)

\(\Rightarrow m_{KMnO_4}=0,8\times158=126,4\left(g\right)\)

Answer 2

\(n_P=\dfrac{12,4}{31}=0,4\left(mol\right)\)
\(n_{O_2}=\dfrac{12,8}{32}=0,4\left(mol\right)\)
a. PTHH: \(4P+5O_2-t^o->2P_2O_5\)
Theo PTHH và đề bài ta có tỉ lệ:
\(\dfrac{0,4}{4}=0,1>\dfrac{0,4}{5}=0,08\)
b. => P dư. \(O_2\) hết => tính theo \(n_{O_2}\)
Theo PT ta có: \(n_{P\left(pư\right)}=\dfrac{0,4.4}{5}=0,32\left(mol\right)\)
=> \(n_{P\left(dư\right)}=0,4-0,32=0,08\left(mol\right)\)
=> \(m_{P\left(dư\right)}=0,08.31=2,48\left(g\right)\)
c. Theo PT ta có: \(n_{P2O5}=\dfrac{0,4.2}{5}=0,16\left(mol\right)\)
=> \(m_{P2O5}=0,16.142=22,72\left(g\right)\)
=> \(m_{sảnphẩmthuđược}=m_{P\left(dư\right)}+m_{P_2O_5}=2,48+22,72=25,2\left(g\right)\)
d. PTHH: \(2KMnO_4-t^o->K_2MnO_4+MnO_2+O_2\uparrow\)
Ta có: \(n_{O_2}=0,4\left(mol\right)\)
Theo PT ta có: \(n_{KMnO_4}=\dfrac{0,4.2}{1}=0,8\left(mol\right)\)
=> \(m_{KMnO_4\left(cầndùng\right)}=0,8.158=126,4\left(g\right)\)