Đốt cháy P của 2,479 lít O2 (đkc)
P+O2------>P2O5
a. tính khối lượng P đã phản ứng
b. tính khối lượng P2O5 sinh ra
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a, PT: \(S+O_2\underrightarrow{t^o}SO_2\)
\(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
b, Ta có: \(n_{SO_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)
Theo PT: \(n_S=n_{SO_2}=0,1\left(mol\right)\)
⇒ mP = 6,3 - mS = 6,3 - 0,1.32 = 3,1 (g)
\(\Rightarrow n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)
Theo PT: \(n_{O_2}=n_S+\dfrac{5}{4}n_P=0,225\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,225.24,79=5,57775\left(l\right)\)
c, Theo PT: \(n_{P_2O_5}=\dfrac{1}{2}n_P=0,05\left(mol\right)\)
\(\Rightarrow m_{P_2O_5}=0,05.142=7,1\left(g\right)\)
a) 4P + 5O2 --to--> 2P2O5
b) \(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)
PTHH:4P + 5O2 --to--> 2P2O5
0,1--------------->0,05
=> \(m_{P_2O_5}=0,05.142=7,1\left(g\right)\)
a, 4P + 5O2 ---> 2P2O5
nP = 3,1/31 = 0,1 (mol )
=> n P2O5 = 0,05 ( mol )
=> m = 0,05 . 142 = 7,1(g)
\(1,PTHH:CaCO_3\underrightarrow{t^o}CaO+CO_2\)
\(áp,dụng.dlbtkl,ta.có:\)
\(m_{CaCO_3}=m_{CaO}+m_{CO_2}\\ m_{CO_2}=m_{CaCO_3}-m_{CaO}=5-2,8=2,2\left(g\right)\)
\(2,a,pthh:4P+5O_2\underrightarrow{t^o}P_2O_5\)
\(n_P=\dfrac{m}{M}=\dfrac{12.4}{31}=0,4\left(mol\right)\)
\(b,theo.pthh\Rightarrow n_{O_2}=\dfrac{5}{4}n_P=\dfrac{5}{4}.0,4=0,5\left(mol\right)\\ \Rightarrow V_{O_2}=n.22,4=0,5.22,4=11,2\left(l\right)\\ m_{O_2}=n.M=0,5.32=16\left(g\right)\)
1. Áp dụng ĐLBTKL, ta có:
\(m_{CaCO_3}=m_{CaO}+m_{CO_2}\)
\(\Leftrightarrow5=2,8+m_{CO_2}\)
\(\Leftrightarrow m_{CO_2}=5-2,8=2,2\left(g\right)\)
2. Ta có: \(n_P=\dfrac{12,4}{31}=0,4\left(mol\right)\)
a. \(PTHH:4P+5O_2\overset{t^o}{--->}2P_2O_5\)
b. Theo PT: \(n_{O_2}=\dfrac{5}{4}.n_P=\dfrac{5}{4}.0,4=0,5\left(mol\right)\)
Theo PT: \(n_{P_2O_5}=\dfrac{1}{2}.n_P=\dfrac{1}{2}.0,4=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}V_{O_2}=0,5.22,4=11,2\left(lít\right)\\m_{P_2O_5}=0,2.142=28,4\left(g\right)\end{matrix}\right.\)
PT: \(S+O_2\underrightarrow{t^o}SO_2\)
\(n_S=\dfrac{6,4}{32}=0,2\left(mol\right)\)
a, \(n_{O_2}=n_S=0,2\left(mol\right)\Rightarrow V_{O_2}=0,2.22,4=4,48\left(l\right)\)
b, \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
\(n_{KMnO_4}=2n_{O_2}=0,4\left(mol\right)\Rightarrow m_{KMnO_4}=0,2.158=31,6\left(g\right)\)
Khối lượng khí Oxi đã phản ứng là:
mP2O5 = mP + mO2
28,4 = 12,4 + mO2
mO2 = 28,4 - 12,4
mO2 = 16 g
\(n_P=\dfrac{m}{M}=\dfrac{6,2}{31}=0,2\left(mol\right)\)
\(PTHH:4P+5O_2-^{t^o}>2P_2O_5\)
tỉ lệ 4 : 5 : 2
n(mol) 0,2---->0,25---->0,1
`V(O_2)=nxx24,79=0,25xx24,79=6,1975(l)`
`V(kk)=6,1975:1/5=30,9875(l)`
a)
\(n_P = \dfrac{6,2}{31} = 0,2(mol)\\ n_{O_2} = \dfrac{6,72}{22,4} = 0,3(mol)\\ 4P + 5O_2 \xrightarrow{t^o} P_2O_5\)
Ta thấy : \(\dfrac{n_P}{4} = 0,05 < \dfrac{n_{O_2}}{5} = 0,06\) nên O2 dư.
Theo PTHH :
\(n_{O_2\ pư} = \dfrac{5}{4}n_P = 0,25(mol)\\ \Rightarrow m_{O_2\ dư} = (0,3-0,25).32 = 1,6(gam)\)
b)
\(n_{P_2O_5} = \dfrac{n_P}{2} = 0,1(mol)\\ \Rightarrow m_{P_2O_5} = 0,1.142 = 14,2(gam)\)
nP= 7,44/31=0,24(mol)
nO2=6,16/22,4=0,275(mol)
PTHH:4 P + 5 O2 -to->2 P2O5
Ta có: 0,24/4 > 0,275/5
=> O2 hết, P dư, tính theo nO2
nP(p.ứ)= 0,275 x 4/5= 0,22(mol)
=>nP(dư)=0,24-0,22=0,02(mol)
=>mP(dư)=0,02.31= 0,62(g)
nP2O5= 2/5 x 0,275= 0,11(mol)
=> mP2O5= 142 x 0,11= 15,62(g)
\(n_P=\dfrac{7,44}{31}=0,24\left(mol\right)\)
\(n_{O_2}=\dfrac{6,16}{22,4}=0,275\left(mol\right)\)
PTHH : \(4P+5O_2\rightarrow2P_2O_5\)
Ban đầu : 0,24 0,275 (mol)
Phản ứng : 0,22 0,275 0,11 (mol)
Sau phản ứng : 0,02 0 0,11 (mol)
\(m_P=0,02.31=0,62\left(g\right)\)
\(m_{P_2O_5}=0,11.142=15,62\left(g\right)\)
Ta có: \(n_{O_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)
PT: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
a, \(n_P=\dfrac{4}{5}n_{O_2}=0,08\left(mol\right)\Rightarrow m_P=0,08.31=2,48\left(g\right)\)
b, \(n_{P_2O_5}=\dfrac{2}{5}n_{O_2}=0,04\left(mol\right)\Rightarrow m_{P_2O_5}=0,04.142=5,68\left(g\right)\)
\(n_{O_2}=\dfrac{2.479}{22.4}=0.1\left(mol\right)\)\(n_{O_2}=\dfrac{2.479}{24.79}=0.1\left(mol\right)\)
\(4P+5O_2\underrightarrow{^{^{t^o}}}2P_2O_5\)
\(0.08....0.1......0.04\)
\(a.m_P=0.08\cdot31=2.48\left(g\right)\)
\(b.m_{P_2O_5}=0.04\cdot142=5.68\left(g\right)\)