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$FeO + CO \xrightarrow{t^o} Fe + CO_2$
Theo PTHH : $n_{FeO} = n_{CO\ pư} = n_{Fe} = n_{CO_2} = a(mol)$
$\Rightarrow m_{giảm} = m_{FeO} - m_{Fe} = 72a -56a = 16a = 1,6(gam)$
$\Rightarrow a = 0,1(mol)$
$m_{Fe} = 0,1.56 = 5,6(gam)$
$n_{CO\ dư} = 0,2 - 0,1 = 0,1(mol)$
$\%V_{CO\ dư} = \%V_{CO_2} = \dfrac{0,1}{0,1 + 0,1}.100\% = 50\%$
nCO phản ứng = nCO2 = nFe = nO = 0,1
→ mFe = 5,6g
nCO2 dư = 0,2 - 0,1 = 0,1
→ %VCO2 = 50%
FeO + CO -> Fe + CO2
nCO=0,2(mol)
Đặt nFeO tham gia PƯ=a
Ta có:
72a-56a=1,6
=>a=0,1
Theo PTHH ta có:
nFe=nCO2=nFeO tham gia PƯ=0,1(mol)
mFe=56.0,1=5,6(g)
%VCO2=\(\dfrac{0,1}{0,2}.100\%=50\%\)
%VCO=100-50=50%
1. n\(_{Ba}\)= \(\dfrac{13,7}{137}\)= 0,1(mol)
n\(O_2\)=\(\dfrac{4,48}{22,4}\)= 0,2(mol)
2Ba+ O\(_2\)\(\rightarrow\)2BaO
Đề bài: 2 1
Pt: 0,1 0,2 (mol)
So sánh: \(\dfrac{n_{Đb}}{n_{Pt}}\)=\(\dfrac{0,1}{2}< \dfrac{0,2}{1}\). Vậy số mol của oxi dư bài toán tính theo số mol của Ba.
\(m_{O_2}\)= 0,2. 32= 6,4(g)
2Ba+ O\(_2\)\(\rightarrow\)2BaO
0,1\(\rightarrow\)0,05 (mol)
\(m_{O_2}\)= 0,05. 32= 1,6(g)
\(m_{O_2}\)(dư)= 6,4-1,6=4,8(g)
3. Đổi: 100(ml)= 0,1(l)
n\(_{Fe}\)=\(\dfrac{5,6}{56}\)= 0,1(mol)
n\(_{HCl}\)= 3.0,1= 0,3(mol)
Fe+ 2HCl\(\rightarrow\)\(FeCl_2\)+ H\(_2\)
Đb: 1 2
Pt: 0,1 0,3 (mol)
S\(^2\): \(\dfrac{n_{Đb}}{n_{Pt}}\)= \(\dfrac{0,1}{1}< \dfrac{0,3}{2}\). Vậy số mol của HCl dư bài toán tính theo số mol của Fe
m\(_{HCl}\)=0,3. 36,5= 10,95(g)
Fe+ 2HCl\(\rightarrow\)\(FeCl_2\)+ H\(_2\)
0,1\(\rightarrow\)0,2 (mol)
m\(_{HCl}\)= 0,2. 36,5= 7,3(g)
m\(_{HCl}\)(dư)= 10,95- 7,3= 3,65(g)
\(2Na+2H_2O\rightarrow2NaOH+H_2\\ Na_2O+H_2O\rightarrow2NaOH\\ n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{Na}=2.0,3=0,6\left(mol\right)\\ a,m_{Na}=0,6.23=13,8\left(g\right)\\ m_{Na_2O}=26,2-13,8=12,4\left(g\right)\\b, n_{Na_2O}=\dfrac{12,4}{62}=0,2\left(mol\right)\\ n_{NaOH\left(tổng\right)}=n_{Na}+2.n_{Na_2O}=0,6+\dfrac{12,4}{62}=0,8\left(mol\right)\\ m_{c.tan}=m_{NaOH}=0,8.40=32\left(g\right)\\ c,m_{ddNaOH}=m_{hh}+m_{H_2O}-m_{H_2}=26,2+200-0,3.2=225,6\left(g\right)\\ C\%_{ddNaOH}=\dfrac{32}{225,6}.100\approx14,185\%\)
PTHH: \(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\uparrow\)
\(Na_2O+H_2O\rightarrow2NaOH\)
Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\) \(\Rightarrow n_{Na}=0,6\left(mol\right)\)
\(\Rightarrow\%m_{Na}=\dfrac{0,6\cdot23}{26,2}\cdot100\%\approx52,67\left(g\right)\) \(\Rightarrow\%m_{Na_2O}=47,33\%\)
Mặt khác: \(n_{Na_2O}=\dfrac{26,2-0,6\cdot23}{62}=0,2\left(mol\right)\)
Theo PTHH: \(n_{NaOH}=n_{Na}+2n_{Na_2O}=1\left(mol\right)\) \(\Rightarrow m_{NaOH}=1\cdot40=40\left(g\right)\)
2Na + 2H2O ---> 2NaOH + H2 (1)
Na2O + H2O ---> 2NaOH (2)
a) nH2 = 0,3 (mol)
Theo pthh (1) : nNa = 2nH2 = 0,6 (mol)
=> mNa = 0,6.23 = 13,8 (g)
=> mNa2O = 26,2 - 13,8 = 12,4 (g)
=> nNa2O = 0,2 (mol)
BTNa : nNaOH = nNa + 2nNa2O = 0,6 + 2.0,2 = 1 (mol)
=> mNaOH = 1.40 = 40(g)
b) %mNa = 13,8.100%/26,2 = 52,67%
%mNa2O = 100% - 52,67% = 47,33%
\(a.n_{Fe_2O_3}=\dfrac{24}{160}=0,15\left(mol\right)\\ n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ Fe_2O_3+3H_2\underrightarrow{to}2Fe+3H_2O\\ Vì:\dfrac{0,3}{3}< \dfrac{0,15}{1}\\ \rightarrow Fe_2O_3dư\\ n_{Fe_2O_3\left(dư\right)}=0,15-\dfrac{0,3}{3}=0,05\left(mol\right)\\ m_{Fe_2O_3\left(dư\right)}=0,05.160=8\left(g\right)\\ b.n_{Fe}=\dfrac{0,3}{3}.2=0,2\left(mol\right)\\ m_{Fe}=0,2.56=11,2\left(g\right)\\ c.m_{rắn}=m_{Fe}+m_{Fe_2O_3\left(dư\right)}=11,2+8=19,2\left(g\right)\)
\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
PTHH: Fe + H2SO4 ---> FeSO4 + H2
0,25<--------------------------0,25
\(\Rightarrow m_{Fe}=0,25.56=14\left(g\right)\\ \Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{14}{32}.100\%=43,75\%\\\%m_{FeO}=100\%-43,75\%=56,25\%\end{matrix}\right.\)
Đặt :
nFeO = x mol
FeO + CO -to-> Fe + CO2
x_____x_______x_____x
m giảm = mFeO - mFe = 1.6
<=> 72x - 56x = 1.6
=> x = 0.1
mFe = 0.1*56 = 5.6 g
nCO dư = 0.2 - 0.1 = 0.1 mol
nCO2 = 0.1 mol
Vì : %V = %n
%CO = %CO2 = 0.1/0.2 *100% = 50%
Đặt \(n_{FeO}=x\left(mol\right)\)
\(FeO+CO\underrightarrow{t^o}Fe+CO_2\)
x → x
\(m_{giảm}=m_{FeO}-m_{Fe}=1,6\)
\(\Leftrightarrow72x-56x=1,6\)
\(\rightarrow x=0,1\)\(\rightarrow m_{Fe}=0,1.56=5,6\left(g\right)\)
\(n_{CO}dư=0,2-0,1=0,1\left(mol\right)\)
→\(n_{CO_2}=0,1\left(mol\right)\)
mà \(\%V=\%n\)
%CO=%CO2\(\frac{0,1}{0,2}\).100%=50%