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a, PTHH: 4P + 5O2 \(\underrightarrow{t^o}\) 2P2O5
b, Theo ĐLBTKL, ta có:
mP + mO\(_2\) = m\(P_2O_5\)
=> mP = 28,4 - 16 = 12,4 (g )
1. \(4P+5O_2\underrightarrow{^{t^o}}2P_2O_5\)
2. Ta có: \(n_P=\dfrac{1,24}{31}=0,04\left(mol\right)\)
Theo PT: \(n_{P_2O_5}=\dfrac{1}{2}n_P=0,02\left(mol\right)\Rightarrow m_{P_2O_5}=0,02.142=2,84\left(g\right)\)
3. \(n_{O_2}=\dfrac{5}{4}n_P=0,05\left(mol\right)\Rightarrow V_{O_2}=0,05.22,4=1,12\left(l\right)\)
\(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\\ n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ PTHH:4P+5O_2\underrightarrow{t^o}2P_2O_5\\ LTL:\dfrac{0,2}{4}< \dfrac{0,4}{5}\Rightarrow O_2dư\)
\(n_{O_2\left(pư\right)}=\dfrac{5}{4}n_P=\dfrac{5}{4}.0,2=0,25\left(mol\right)\\ n_{O_2\left(dư\right)}=0,4-0,25=0,15\left(mol\right)\)
\(n_{P_2O_5\left(lt\right)}=\dfrac{1}{2}n_P=\dfrac{1}{2}.0,2=0,1\left(mol\right)\\ m_{P_2O_5\left(lt\right)}=0,1.142=14,2\left(g\right)\\ m_{P_2O_5\left(tt\right)}=0,1.142.80\%=11,36\left(g\right)\)
\(PTHH:4P+5O_2\rightarrow2P_2O_5\)
a) \(n_P=\frac{6,2}{31}=0,2\left(mol\right)\)
Theo PTHH:
\(n_{P2O5}=0,5.n_P=0,5.0,2=0,1\left(mol\right)\)
\(m_{P2O5}=0,1.142=14,2\left(g\right)\)
b) \(n_{O2}=\frac{11,2}{22,4}=0,5\left(mol\right)\)
Theo PTHH:
\(n_{P2O5}=\frac{2}{5}.n_{O2}=0,2\left(mol\right)\)
\(m_{P2O5}=142.0,2=28,4\left(g\right)\)
c) \(n_P=12,431=0,4\left(mol\right)\)
\(n_{O2}=\frac{17}{32}=0,53125\left(mol\right)\)
Lập tỉ lệ: \(\frac{0,4}{4}< \frac{0,53125}{5}\)
Nên O2 dư, P hết
Theo PTHH:
\(n_{P2O5}=0,5n_P=0,5.0,4=0,2\left(mol\right)\)
\(m_{P2O5}=142.0,2=28,4\left(g\right)\)
e 1 )
\(n_{O2\left(dư\right)}=0,53125-0,5=0,03125\left(mol\right)\)
\(V_{O2\left(Dư\right)}=22,4.0,03125=0,7\left(mol\right)\)
d) \(n_P=\frac{15,5}{31}=0,5\left(mol\right)\)
\(n_{O2}=\frac{11,2}{22,4}=0,5\left(mol\right)\)
Tỉ lệ: \(\frac{0,5}{4}>\frac{0,5}{5}\)
Nên P dư, O2 hết
Theo PTHH: \(n_{P2O5}=\frac{2}{5}n_{O2}=0,2\left(mol\right)\)
\(\Rightarrow m_{P2O5}=142.0,2=28,4\left(g\right)\)
e2)
\(n_{P\left(dư\right)}=0,5.\left(\frac{4}{5}.0,5\right)=0,1\left(mol\right)\)
\(m_{P\left(dư\right)}=31.0,1=3,1\left(g\right)\)
a) PTHH: 4P + 5O2 → 2P2O5
Sản phẩm thuộc oxit axit.
b) nP = \(\frac{m_P}{M_P}=\frac{1,24}{31}=0,04\left(mol\right)\)
Theo PTHH: Cứ 4 mol P phản ứng thì tạo ra 2 mol P2O5
=> Cứ 0,04 mol P phản ứng thì tạo ra 0,02 mol P2O5
=> mP2O5 = n.M = 0,02 . 142 = 2,84(g)
c) nO2 = \(\frac{V_{O2\left(\text{đ}ktc\right)}}{22,4}=\frac{13,44}{22,4}=0,6\left(mol\right)\)
(So sánh tỉ số: \(\frac{0,04}{4}< \frac{0,6}{5}\) => khí O2 dư)
Ta có: 4P + 5O2 → 2P2O5
Ban đầu: 0,04 0,6 (mol)
P/ứng: 0,04 0,05 0,02 (mol)
Sau p/ứng: 0 0,55 0,02 (mol)
=> Khối lượng khí oxi dư là:
mO2 = n.M = 0,55 . 32 = 17,6 (g)
a)\(4P+5O2--->2P2O5\)
Sản phẩm là oxit axit
b)\(n_P=\frac{1,24}{31}=0,04\left(mol\right)\)
\(n_{P2O5}=\frac{1}{2}n_P=0,02\left(mol\right)\)
\(m_{P2O5}=0,02.142=2,84\left(g\right)\)
c)\(n_{O2}=\frac{13,44}{22,4}=0,6\left(mol\right)\)
\(n_P\left(\frac{0,04}{4}\right)< n_{O2}\left(\frac{0,6}{5}\right)\)
=>O2 dư
\(n_{O2}=\frac{5}{4}n_P=0,05\left(mol\right)\)
\(n_{O2}dư=0,6-0,05=0,45\left(mol\right)\)
\(m_{O2}dư=0,45.32=14,4\left(g\right)\)
a) 4P + 5O2 --to--> 2P2O5
b) \(n_{P_2O_5}=\dfrac{34,08}{142}=0,24\left(mol\right)\)
4P + 5O2 --to--> 2P2O5
0,48<-0,6<------0,24
=> mO2 = 0,6.32 = 19,2 (g)
c)
C1: mP = 0,48.31 = 14,88(g)
C2:
Theo ĐLBTKL: mP + mO2 = mP2O5
=> mP = 34,08-19,2 = 14,88(g)
d)
VO2 = 0,6.22,4 = 13,44 (l)
=> Vkk = 13,44 :20% = 67,2 (l)
a, PT: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
b, Ta có: \(n_{P_2O_5}=\dfrac{7,1}{142}=0,05\left(mol\right)\)
Theo PT: \(n_P=2n_{P_2O_5}=0,1\left(mol\right)\)
\(\Rightarrow m_P=0,1.31=3,1\left(g\right)\)
\(n_{O_2}=\dfrac{5}{2}n_{P_2O_5}=0,125\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,125.22,4=2,8\left(l\right)\)
c, Có: \(V_{O_2\left(dư\right)}=2,8.15\%=0,42\left(l\right)\)
\(\Rightarrow V_{O_2}=2,8+0,42=3,22\left(l\right)\)
\(n_{P2O5}=\frac{35,5}{142}=0,25\left(mol\right)\)
\(4P+5O_2\underrightarrow{t^O}2P_2O_5\)
0,5___0,625___0,25(mol)
\(x=0,5.31=15,5\left(g\right)\)
\(y=0,625.22,4=14\left(l\right)\)
\(\Rightarrow ChọnđápánC\)