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Ta có: \(n_{Na}=\dfrac{2,3}{23}=0,1\left(mol\right)\)
a. PTHH: 2Na + 2H2O ---> 2NaOH + H2↑
b. Ta có: \(n_{H_2O}=\dfrac{97,8}{18}=5,43\left(mol\right)\)
Ta thấy: \(\dfrac{0,1}{2}< \dfrac{5,43}{2}\)
=> H2O dư.
Theo PT: \(n_{H_2}=\dfrac{1}{2}.n_{Na}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\)
=> \(V_{H_2}=0,05.22,4=1,12\left(lít\right)\)
c. Ta có: \(m_{dd_{NaOH}}=2,3+97,8=100,1\left(g\right)\)
Theo PT: \(n_{NaOH}=n_{Na}=0,1\left(mol\right)\)
=> \(m_{NaOH}=0,1.40=4\left(g\right)\)
=> \(C_{\%_{NaOH}}=\dfrac{4}{100,1}.100\%=3,996\%\)
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, \(2Na+2H_2O\rightarrow2NaOH+H_2\)
Ta có: \(n_{Na}=\dfrac{2,3}{23}=0,1\left(mol\right)\)
\(n_{H_2O}=\dfrac{200}{18}=\dfrac{100}{9}\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,1}{2}< \dfrac{\dfrac{100}{9}}{2}\), ta được H2O dư.
Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{Na}=0,05\left(mol\right)\Rightarrow V_{H_2}=0,05.22,4=1,12\left(l\right)\)
b, Theo PT: \(n_{NaOH}=n_{Na}=0,1\left(mol\right)\)
Ta có: m dd sau pư = 2,3 + 100 - 0,05.2 = 102,2 (g)
\(\Rightarrow C\%_{NaOH}=\dfrac{0,1.40}{102,2}.100\%\approx3,91\%\)
c, - Dung dịch làm quỳ tím hóa xanh.
\(n_{Na}=\dfrac{2,3}{23}=0,1\left(mol\right)\\ 2Na+2H_2O\rightarrow2NaOH+H_2\\ n_{H_2}=\dfrac{0,1}{2}=0,05\left(mol\right);n_{NaOH}=n_{Na}=0,1\left(mol\right)\\ a,V=V_{H_2\left(đktc\right)}=0,05.22,4=1,12\left(l\right)\\ b,m_{ddNaOH}=m_{Na}+m_{H_2O}-m_{H_2}=2,3+200-0,05.2=202,2\left(g\right)\\ C\%_{ddNaOH}=\dfrac{40.0,1}{202,2}.100\approx1,978\%\\ c,NaOH-Tính.bazo\Rightarrow Quỳ.tím.hoá.xanh\)
\(a) 4Na + O_2 \xrightarrow{t^o} 2Na_2O\\ b) n_{Na} = \dfrac{4,6}{23} = 0,2(mol)\\ n_{O_2} = \dfrac{1}{4}n_{Na} = 0,05(mol)\\ V_{O_2} = 0,05.22,4 = 1,12(lít)\\ c) Na_2O + H_2O \to 2NaOH\\ n_{NaOH} = n_{Na} = 0,2(mol)\\ C\%_{NaOH} = \dfrac{0,2.40}{160}.100\% = 5\%\\ d)\)
\(n_{Na\ thêm} = x(mol)\\ 2Na + 2H_2O \to 2NaOH + H_2\\ n_{NaOH} = n_{Na} = x(mol)\\ n_{H_2} =0,5x(mol)\\ \Rightarrow m_{dd} = 23x + 160 -0,5x.2 = 22x + 160(gam)\\ \Rightarrow C\% = \dfrac{0,2.40 + 40x}{22x + 160}.100\% = 5\% + 5\%\\ \Rightarrow x = \dfrac{40}{189}\\ m_{Na} = \dfrac{40}{189}.23 = 4,87(gam)\)
\(n_{Na}=\dfrac{4.6}{23}=0.2\left(mol\right)\)
\(4Na+O_2\underrightarrow{^{t^0}}2Na_2O\)
\(0.2.....0.05.........0.1\)
\(V_{O_2}=0.05\cdot22.4=1.12\left(l\right)\)
\(Na_2O+H_2O\rightarrow2NaOH\)
\(0.1.......................0.2\)
\(m_{NaOH}=0.2\cdot40=8\left(g\right)\)
\(C\%_{NaOH}=\dfrac{8}{160}\cdot100\%=5\%\)
Để C% tăng thêm 5%
\(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)
\(a...............a.......0.5a\)
\(m_{NaOH}=40a\left(g\right)\)
\(m_{dd_{NaOH}}=23a+160-0.5a\cdot2=22a+160\left(g\right)\)
\(C\%_{NaOH}=\dfrac{40a+8}{22a+160}\cdot100\%=5\%\)
\(\Rightarrow a=0\)
=> Sai đề
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%