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nP = 6.2/31 = 0.2 (mol)
nO2 = 6.72/22.4 = 0.3 (mol)
4P + 5O2 -to-> 2P2O5
0.2___0.25_____0.1
mO2 dư = ( 0.3 - 0.25) * 32 = 1.6(g)
mP2O5 = 0.1*142 = 14.2 (g)
Ta có: \(n_P=\dfrac{6.2}{31}=0.29mol\)
\(n_{O_2}=\dfrac{6.72}{22.4}=0.3mol\)
PTHH:
\(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
ta có:
\(\left\{{}\begin{matrix}\dfrac{n_{P\left(bra\right)}}{nP_{\left(pthh\right)}}=\dfrac{0.2}{4}=0.05\\\dfrac{n_{O_2\left(bra\right)}}{n_{O_2}\left(pthh\right)}=\dfrac{0.3}{5}=0.06\end{matrix}\right.\)
=> \(O_2\) dư
\(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
4 ----------->2
0.2---------->0.1=nP2O5
=>\(m_{P_2O_5}=142.0.1=14.2\left(g\right)\)
a, \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
b, \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)
\(n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,2}{4}< \dfrac{0,4}{5}\), ta được O2 dư.
Theo PT: \(n_{O_2\left(pư\right)}=\dfrac{5}{4}n_P=0,25\left(mol\right)\Rightarrow n_{O_2\left(dư\right)}=0,4-0,25=0,15\left(mol\right)\)
\(\Rightarrow m_{O_2\left(dư\right)}=0,15.32=4,8\left(g\right)\)
c, Theo PT: \(n_{P_2O_5}=\dfrac{1}{2}n_P=0,1\left(mol\right)\Rightarrow m_{P_2O_5}=0,1.142=14,2\left(g\right)\)
d, \(m_{P_2O_5}=14,2.80\%=11,36\left(g\right)\)
\(a) n_P = \dfrac{6,2}{31} = 0,2(mol) ;n_{O_2} = \dfrac{6,72}{22,4} = 0,3(mol)\\ 4P + 5O_2 \xrightarrow{t^o} 2P_2O_5\\ \dfrac{n_P}{4} = 0,05< \dfrac{n_{O_2}}{5} = 0,06 \to O_2\ dư\\ n_{O_2\ pư} = \dfrac{5}{4}n_P = 0,25(mol)\\ m_{O_2\ dư} = (0,3 -0,25).32 = 1,6(gam)\\ b) n_{P_2O_5} = \dfrac{1}{2}n_P = 0,1(mol) \Rightarrow m_{P_2O_5} = 0,1.142 = 14,2(gam)\)
\(a.4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\\ n_P=\dfrac{6,2}{31}=0,2\left(mol\right);n_{O_2}=\dfrac{19,2}{32}=0,6\left(mol\right)\\ Vì:\dfrac{0,2}{4}< \dfrac{0,6}{5}\\ \Rightarrow O_2dư\\ \Rightarrow n_{O_2\left(dư\right)}=0,6-\dfrac{5}{4}.0,2=0,35\left(mol\right)\\ \Rightarrow m_{O_2\left(dư\right)}=0,35.32=11,2\left(g\right)\\ b,n_{P_2O_5}=\dfrac{n_P}{2}=\dfrac{0,2}{2}=0,1\left(mol\right)\\ \Rightarrow m_{P_2O_5}=142.0,1=14,2\left(g\right)\)
Tham khảo
nP = 6.2/31 = 0.2 (mol)
nO2 = 6.72/22.4 = 0.3 (mol)
4P + 5O2 -to-> 2P2O5
0.2___0.25_____0.1
mO2 dư = ( 0.3 - 0.25) * 32 = 1.6(g)
mP2O5 = 0.1*142 = 14.2 (g)
Bài 1:
\(a,2Cu+O_2\underrightarrow{t^o}2CuO\)
b, \(n_{O_2}=\dfrac{1,12}{32}=0,035mol\)
\(n_{Cu}=\dfrac{6,4}{64}=0,1mol\)
\(\dfrac{0,1}{2}>\dfrac{0,035}{1}\) => Cu dư, O2 đủ
\(n_{Cu}\left(dư\right)=0,1-0,07=0,039\left(mol\right)\)
c, \(m_{CuO}=0,07.80=5,6g\)
Bài 2:
\(n_{Al}=\dfrac{13,5}{27}=0,5mol\)
\(n_{O_2}=\dfrac{6,67}{32}=0,21\left(mol\right)\)
\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
\(\dfrac{0,5}{4}>\dfrac{0,21}{3}\) => Al dư, O2 đủ
\(n_{Al_2O_3}=\dfrac{2}{3}.0,21=0,14\left(mol\right)\)
\(m_{Al_2O_3}=0,14.102=14,28g\)
nP=\(\dfrac{62}{31}\)=0,2(mol)
nO2=\(\dfrac{7,84}{22,4}\)=0,35(mol)
PTHH:4P+5O2to→2P2O5
tpứ: 0,2 0,35
pứ: 0,2 0,25 0,1
spứ: 0 0,1 0,1
a)chất còn dư là oxi
mO2dư=0,1.32=3,2(g)
b)mP2O5=n.M=0,1.142=14,2(g)
\(a.n_P=0,2\left(mol\right);n_{O_2}=0,35\left(mol\right)\\ 4P+5O_2-^{t^o}\rightarrow2P_2O_5\\ LTL:\dfrac{0,2}{4}< \dfrac{0,35}{5}\\ \Rightarrow SauphảnứngO_2dư\\ n_{O_2\left(pứ\right)}=\dfrac{5}{4}n_P=0,25\left(mol\right)\\ \Rightarrow m_{P\left(dư\right)}=\left(0,35-0,25\right).32=3,2\left(g\right)\\ b.n_{P_2O_5}=\dfrac{1}{2}n_P=0,1\left(mol\right)\\ \Rightarrow m_{P_2O_5}=0,1.142=14,2\left(g\right)\)
\(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\\ n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ PTHH:4P+5O_2\underrightarrow{t^o}2P_2O_5\\ LTL:\dfrac{0,2}{4}< \dfrac{0,4}{5}\Rightarrow O_2dư\)
\(n_{O_2\left(pư\right)}=\dfrac{5}{4}n_P=\dfrac{5}{4}.0,2=0,25\left(mol\right)\\ n_{O_2\left(dư\right)}=0,4-0,25=0,15\left(mol\right)\)
\(n_{P_2O_5\left(lt\right)}=\dfrac{1}{2}n_P=\dfrac{1}{2}.0,2=0,1\left(mol\right)\\ m_{P_2O_5\left(lt\right)}=0,1.142=14,2\left(g\right)\\ m_{P_2O_5\left(tt\right)}=0,1.142.80\%=11,36\left(g\right)\)
a) nZn = 26/65 = 0,4 (mol)
PTHH: 2Zn + O2 -> (t°) 2ZnO
LTL: 0,4/2 < 0,3 => O2 dư
nO2 (p/ư) = 0,4/2 = 0,2 (mol)
mO2 (dư) = (0,3 - 0,2) . 32 = 3,2 (g(
b) nZnO = 0,4 (mol)
mZnO = 0,4 . 81 = 32,4 (g)
a. \(n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\)
\(n_{O_2}=\dfrac{7.437}{24,79}=0,3\left(mol\right)\)
Ta thấy : 0,4 > 0,3 => Zn dư , O2 đủ
PTHH : 2Zn + O2 ----to---> 2ZnO
0,6 0,3 0,6
\(m_{Zn\left(dư\right)}=\left(0,4-0,6\right).65=-13\left(g\right)\)
b. \(m_{ZnO}=0,6.81=48,6\left(g\right)\)
C9:
nP = 6,2/31 = 0,2 (mol)
nO2 = 6,4/32 = 0,2 (mol)
PTHH: 4P + 5O2 -> (t°) 2P2O5
LTL: 0,2/4 > 0,2/5 => P dư
nP (p/ư) = 0,2/5 . 4 = 0,16 (mol)
nP (dư) = 0,2 - 0,16 = 0,04 (mol)
nP2O5 = 0,2/5 . 2 = 0,08 (mol)
mP2O5 = 0,08 . 142 = 11,36 (g)
C10:
Áp dụng ĐLBTKL, ta có:
mR + mO2 = mRO
=> mO2 = 21,6 - 16,8 = 4,8 (g(
=> nO2 = 4,8/32 = 0,15 (mol)
PTHH: 2R + O2 -> (t°) 2RO
nR = 0,15 . 2 = 0,3 (mol)
M(R) = 16,8/0,3 = 56 (g/mol(
=> R là Fe
a)
\(n_P = \dfrac{6,2}{31} = 0,2(mol) ; n_{O_2} = \dfrac{6,72}{22,4} = 0,3(mol)\\ 4P + 5O_2 \xrightarrow{t^o} 2P_2O_5\)
Ta thấy :
\(\dfrac{n_P}{4} = 0,05 < \dfrac{n_{O_2}}{5} = 0,06\) nên O2 dư.
\(n_{O_2\ pư} = \dfrac{5}{4}n_P = 0,25(mol)\\ \Rightarrow m_{O_2\ dư} = (0,3 - 0,25).32 = 1,6(gam)\)
b)
\(n_{P_2O_5} = \dfrac{n_P}{2} = 0,1(mol)$\\ m_{P_2O_5} = 0,1.142 = 14,2(gam)\)