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a, PTHH:
(1) 2Na + 2H2O→ 2NaOH + H2
2mol 2mol 2mol 1mol
(2) Na2O + H2O → 2NaOH
1mol 1mol 2mol
b,
-Tính nH2
nH2= \(\dfrac{3.36}{22.4}\)=0.15 → mH2= 0.15*2=0.3(g)
-Theo PT(1) ta có
nNa=nNaOH(1)=2nH2= 2*0.15=0.3(mol)
→mNa= 0.3*23= 6.9(g)
mNaOH(1)= 0.3*40=12(g)
-Tính mNa2O
mNa2O= 13.1-6.9= 6.2(g)
→nNa2O= \(\dfrac{6.2}{62}\)=0.1(mol)
-Theo PT(2) ta có
nNaOH(2)=2Na2O=2*0.1=0.2(mol)
→mNaOH(2)=0.2*23=4.6(g)
-Tính mNaOH
mNaOH= mNaOH(1)+mNaOH(2)=12+4.6=16.6(g)
-Tính C%
C%= \(\dfrac{16.6}{100+13.1-0.3}\)*100%≃ 14.7%
c,
PTHH3:CO2 +2NaOH → Na2CO3 + H2O
1mol 2mol 1mol 1mol
-Theo Đb ta có:
mNaOH(3) = 1/2mNaOH(1)+(2)= \(\dfrac{16.6}{2}\)= 8.3(g)
→nNaOH(3)= \(\dfrac{8.3}{40}\)= 0.2075(mol)
-TheoPT(3) ta có
nNa2CO3=1/2nNaOH(3)= \(\dfrac{0.2075}{2}\)=0.10375(mol)
→mNa2CO3=0.10375*106≃11(g)
-Tính C%
C%=\(\dfrac{11}{11+1.8675}\)*100%≃ 85.4867%
nNa2CO3= 25/106(mol)
PTHH: Na2CO3 + 2 HCl -> 2 NaCl + CO2 + H2O
a) nHCl=25/106 . 2= 25/53 (mol)
=> m=mddHCl={[25/53].36,5]/15%}=114,78(g)
b) nCO2= 25/106 x 22,4= 5,28(l)
c) mNaCl=25/53. 58,5=27,59(g)
mddNaCl=25+114,78- 25/106.44=129,4(g)
=>C%ddNaCl=(27,59/129,4).100=21,32%
a) \(\left\{{}\begin{matrix}n_{CH_4}=\dfrac{1,68}{22,4}=0,075\left(mol\right)\\n_{Na}=\dfrac{2,3}{23}=0,1\left(mol\right)\end{matrix}\right.\)
PTHH:
\(CH_4+2O_2\xrightarrow[]{t^o}CO_2+2H_2O\)
0,075--------->0,075
\(2Na+2H_2O\rightarrow2NaOH+H_2\)
0,1---------------->0,1
b) \(m_{NaOH}=0,1.40=4\left(g\right)\)
c) Xét \(T=\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{0,1}{0,075}=\dfrac{4}{3}\)
\(1< \dfrac{4}{3}< 2\Rightarrow\) Pư tạo 2 muối
Đặt \(\left\{{}\begin{matrix}n_{Na_2CO_3}=a\left(mol\right)\\n_{NaHCO_3}=b\left(mol\right)\end{matrix}\right.\)
PTHH:
\(2NaOH+CO_2\rightarrow Na_2CO_3+H_2O\)
2a<---------a<-------a
\(NaOH+CO_2\rightarrow NaHCO_3\)
b<----------b<-------b
\(\Rightarrow\left\{{}\begin{matrix}2a+b=0,1\\a+b=0,075\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,025\\b=0,05\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}C_{M\left(Na_2CO_3\right)}=\dfrac{0,025}{0,25}=0,1M\\C_{M\left(NaHCO_3\right)}=\dfrac{0,05}{0,25}=0,2M\end{matrix}\right.\)
\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
pt : \(Fe+2HCl\rightarrow FeCl_2+H_2|\)
1 2 1 1
0,2 0,4 0,2 0,2
a) \(n_{HCl}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)
\(m_{HCl}=0,4.36,5=14,6\left(g\right)\)
\(m_{ddHCl}=\dfrac{14,6.100}{14,6}=100\left(g\right)\)
\(n_{H2}=\dfrac{0,4.1}{2}=0,2\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,2.22,4=4,48\left(l\right)\)
b) \(n_{FeCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
⇒ \(m_{FeCl2}=0,2.127=25,4\left(g\right)\)
\(m_{ddspu}=11,2+100-\left(0,2.2\right)=110,8\left(g\right)\)
\(C_{FeCl2}=\dfrac{25,4.100}{110,8}=22,92\)0/0
Chúc bạn học tốt
\(n_K=\dfrac{5,85}{39}=0,15\left(mol\right)\)
PTHH: 2K + 2H2O --> 2KOH + H2
_____0,15------------->0,15-->0,075
=> VH2 = 0,075.22,4 =1,68(l)
mdd = 5,85 + 100 - 0,075.2 = 105,7(g)
=> \(C\%=\dfrac{0,15.56}{105,7}.100\%=7,95\%\)
Na tác dụng với axit trước.
Dung dịch sau phản ứng gồm: Na+; Cl-; SO42-
⟶m = 28,952
Đáp án D
Bài 1: Ta có: \(n_{H_2}=\dfrac{1,008}{22,4}=0,045\left(mol\right)\)
PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
__0,045__0,09____0,045___0,045 (mol)
a, Ta có: \(a=m_{Mg}=0,045.24=1,08\left(g\right)\)
b, \(V_{ddHCl}=\dfrac{0,09}{0,1}=0,9\left(l\right)\)
c, \(C_{M_{MgCl_2}}=\dfrac{0,045}{0,9}=0,05M\)
Bài 2:
PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
a, Giả sử: \(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)
⇒ 24x + 56y = 5,2 (1)
Ta có: \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Mg}+n_{Fe}=x+y\left(mol\right)\)
⇒ x + y = 0,15 (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,1\left(mol\right)\\y=0,05\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,1.24}{5,2}.100\%\approx46,2\%\\\%m_{Fe}\approx53,8\%\end{matrix}\right.\)
b, Ta có: \(n_{HCl}=2n_{H_2}=0,3\left(mol\right)\)
\(\Rightarrow V_{HCl}=\dfrac{0,3}{1}=0,3\left(l\right)\)
Bạn tham khảo nhé!
