Question

Đốt cháy hết 6,2g photpho. Hãy tính:

1. Khối lượng diphotpho pentaoxit-P₂O₅ thu dc

2. Khối lượng khí oxi đã dùng bằng 2 cách

(P= 32; O=16)

Answer 1

Answer 2

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

\(n_P=\frac{6,2}{31}=0,2\left(mol\right)\)

1) Theo pt: \(n_{P_2O_5}=\frac{1}{2}n_P=\frac{1}{2}\times0,2=0,1\left(mol\right)\)

\(\Rightarrow m_{P_2O_5}=0,1\times142=14,2\left(g\right)\)

2) C1: Theo pT: \(n_{O_2}=\frac{5}{4}n_P=\frac{5}{4}\times0,2=0,25\left(mol\right)\)

\(\Rightarrow m_{O_2}=0,25\times32=8\left(g\right)\)

C2: Theo ĐL BTKL ta có:

\(m_P+m_{O_2}=m_{P_2O_5}\)

\(\Leftrightarrow m_{O_2}=m_{P_2O_5}-m_P=14,2-6,2=8\left(g\right)\)

Answer 3

4P + 5O₂ --> 2P₂O₅

1. n_P = \(\frac{6,2}{31}\) = 0,2 mol

Theo PTHH, ta có:

n_{\(P_2O_5\)} = 0,1 mol

m_{\(P_2O_5\)} = 0,1.142 = 14,2 g

2.

C1: Theo ĐLBTKL, ta có:

m_{\(O_2\)} + \(m_P\) = m_{\(P_2O_5\)}

=> m_{\(O_2\)} = m_{\(P_2O_5\)} - \(m_P\) = 14,2 - 6,2 = 8g

C2:Theo PTHH, ta có: n_{\(O_2\)} = 0,25 mol

=> m_{\(O_2\)} = 0,25 . 32 =8 g

Answer 4

\(n_P=\frac{m}{M}=\frac{6,2}{31}=0,2\left(mol\right)\\ PTHH:4P+5O_2\underrightarrow{t^o}2P_2O_5\)

(mol) 4 5 2

(mol) 0,2 0,25 0,1

\(m_{P_2O_5}=n.M=0,1.142=14,2\left(g\right)\\ C_1:m_{O_2}=n.M=0,25.32=8\left(g\right)\\ C_2:Theo.DLBTKL\\ \Rightarrow M_{O_2}=m_{P_2O_5}-m_P=14,2-6,2=8\left(g\right)\)

Answer 5

1) \(n_P=\frac{6,2}{31}=0,2\left(mol\right)\)

PTHH: \(4P+5O_2\underrightarrow{t^0}2P_2O_5\)

Theo PTHH: \(n_P:n_{P_2O_5}=4:2=2:1\)

\(\Rightarrow n_{P_2O_5}=n_P.\frac{1}{2}=0,2.\frac{1}{2}=0,1\left(mol\right)\)

\(\Rightarrow m_{P_2O_5}=0,1.142=14,2\left(g\right)\)

2) Cách 1: Theo PTHH: \(n_P:n_{O_2}=4:5\)

\(\Rightarrow n_{O_2}=n_P.\frac{5}{4}=0,2.\frac{5}{4}=0,25\left(mol\right)\)

\(\Rightarrow m_{O_2}=0,25.32=8\left(g\right)\)

Cách 2: Áp dụng định luật bảo toàn khối lượng có:

\(m_P+m_{O_2}=m_{P_2O_5}\)

\(\Rightarrow m_{O_2}=m_{P_2O_5}-m_P=14,2-6,2=8\left(g\right)\)