Câu 13: Cần cho bao nhiêu gam Na2O vào 200g dung dịch NaOH 16% để thu được dung dịch mới có nồng độ 20%
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\(n_{Na}=\dfrac{23}{23}=1\left(mol\right)\)
PTHH: 2Na + 2H2O --> 2NaOH + H2
1------------------->1---->0,5
=> mNaOH = 1.40 = 40 (g)
mdd sau pư = 23 + 100 - 0,5.2 = 122 (g)
=> \(C\%=\dfrac{40}{122}.100\%=32,787\%\)
a) 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
b) \(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)
PTHH: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
0,4--->0,6-------------------->0,6
=> VH2 = 0,6.22,4 = 13,44 (l)
c) \(V_{dd.H_2SO_4}=\dfrac{0,6}{1}=0,6\left(l\right)\)
d) \(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)
PTHH: Fe2O3 + 3H2 --to--> 2Fe + 3H2O
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,6}{3}\) => Fe2O3 hết, H2 dư
PTHH: Fe2O3 + 3H2 --to--> 2Fe + 3H2O
0,1----------------->0,2
=> mFe = 0,2.56 = 11,2 (g)
\(n_{SO_3}=\dfrac{100}{80}=1,25\left(mol\right)\\ n_{H_2SO_4\left(bđ\right)}=1.3=3\left(mol\right)\\ m_{H_2SO_4\left(bđ\right)}=3.98=294\left(g\right)\\ m_{ddH_2SO_4\left(bđ\right)}=1000.1.1,12=1120\left(g\right)\\ m_{ddH_2SO_4\left(sau\right)}=1120+100=1220\left(g\right)\)
PTHH: SO3 + H2O ---> H2SO4
1,25-------------->1,25
=> \(m_{H_2SO_4\left(sau\right)}=294+1,25.98=416,5\left(g\right)\)
=> \(C\%_{H_2SO_4\left(sau\right)}=\dfrac{416,5}{1220}.100\%=34,14\%\)
1)
a) Fe + 2HCl --> FeCl2 + H2
b) \(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
0,15->0,3--->0,15-->0,15
=> VH2 = 0,15.22,4 = 3,36 (l)
c) mdd sau pư = 8,4 + 250 - 0,15.2 = 258,1 (g)
=> \(C\%_{FeCl_2}=\dfrac{0,15.127}{258,1}.100\%=7,38\%\)
2)
a) Zn + 2HCl --> ZnCl2 + H2
b) \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
0,2-->0,4---->0,2--->0,2
=> VH2 = 0,2.22,4 = 4,48 (l)
mZnCl2 = 0,2.136 = 27,2 (g)
c) \(C_{M\left(dd.HCl\right)}=\dfrac{0,4}{0,2}=2M\)
d)
PTHH: A + 2HCl --> ACl2 + H2
0,2<--0,4
=> \(M_A=\dfrac{4,8}{0,2}=24\left(g/mol\right)\)
=> A là Mg(Magie)
1
\(4Na+O_2\underrightarrow{t^o}2Na_2O\\
Na_2O+H_2O\rightarrow2NaOH\\
b,4P+5O_2\underrightarrow{t^o}2P_2O_5\\
P_2O_5+3H_2O\underrightarrow{t^o}2H_3PO_4\\
c,2Cu+O_2\underrightarrow{t^o}2CuO\\
CuO+H_2O\underrightarrow{t^o}Cu+H_2O\)
2
\(Na+H_2O\rightarrow NaCl+\dfrac{1}{2}H_2\\
CaO+H_2O\rightarrow Ca\left(OH\right)_2\\
SO_2+H_2O\rightarrow H_2SO_3\\
P_2O_5+3H_2O\rightarrow2H_3PO_4\\
Ba+H_2O\rightarrow Ba\left(OH\right)_2\\
b,Zn+2HCl\rightarrow ZnCl_2+H_2\\
Na+2HCl\rightarrow2NaCl+H_2\)
1
4Na+O2to→2Na2ONa2O+H2O→2NaOHb,4P+5O2to→2P2O5P2O5+3H2Oto→2H3PO4c,2Cu+O2to→2CuO
\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\
pthh:Fe+2HCl\rightarrow FeCl_2+H_2\)
0,2 0,4 0,2
\(V_{H_2}=0,2.22,4=4,48l\\
V_{HCl}=\dfrac{0,4}{1}=0,4\left(l\right)\)
2)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\
pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,2 0,4 0,2
\(V_{H_2}=0,222,4=4,48l\\
C\%_{HCl}=\dfrac{0,4.36,5}{200}.100\%=7,3\%\)
\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\
pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,1 0,2 0,1
\(V_{H_2}=0,1.22,4=2,24l\\
m_{HCl}=\left(0,2.36,5\right).10\%=0,73g\)
\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\\
pthh:Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\\
LTL:\dfrac{0,1}{1}>\dfrac{0,1}{3}\)
=> Fe2O3 dư
\(n_{Fe}=\dfrac{2}{3}n_{H_2}=0,067\left(mol\right)\\
m_{Fe}=0,067.56=3,73g\)
a.b.\(n_{Zn}=\dfrac{6,5}{65}=0,1mol\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,1 0,2 0,1 ( mol )
\(V_{H_2}=0,1.22,4=2,24l\)
\(m_{ddHCl}=\dfrac{0,2.36,5}{10\%}=73g\)
c.\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1mol\)
\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)
0,1 > 0,1 ( mol )
0,1 1/15 ( mol )
\(m_{Fe}=\dfrac{1}{15}.56=3,73g\)
Gọi số mol Na2O cần thêm là a (mol)
\(m_{NaOH\left(bđ\right)}=\dfrac{200.16}{100}=32\left(g\right)\)
PTHH: Na2O + H2O --> 2NaOH
a---------------->2a
=> mNaOH(dd sau pư) = 40.2a + 32 = 80a + 32 (g)
mdd sau pư = 62a + 200 (g)
=> \(C\%=\dfrac{80a+32}{62a+200}.100\%=20\%\)
=> \(a=\dfrac{20}{169}\left(mol\right)\Rightarrow n_{Na_2O}=\dfrac{20}{169}.62=\dfrac{1240}{169}\left(g\right)\)