Cho 15,5g Na2O tác dụng với 15,5 lít đ HCl a) Tính khối lượng muooiz tạo thành b) tính nồng độc mol của dd axit đã dùng
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a, \(Na_2O+H_2O\rightarrow2NaOH\)
Ta có: \(n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\)
Theo PT: \(n_{NaOH}=2n_{Na_2O}=0,5\left(mol\right)\)
\(\Rightarrow CM_{NaOH}=\dfrac{0,5}{0,5}=1\left(M\right)\)
b, \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
Theo PT: \(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,25\left(mol\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,25.98}{20\%}=122,5\left(g\right)\)
\(\Rightarrow V_{ddH_2SO_4}=\dfrac{122,5}{1,14}\approx107,46\left(ml\right)\)
a. PTHH: AgNO3 + HCl ---> AgCl↓ + HNO3
b. Ta có: \(n_{AgNO_3}=\dfrac{42,5}{170}=0,25\left(mol\right)\)
Theo PT: \(n_{AgCl}=n_{AgNO_3}=0,25\left(mol\right)\)
=> \(m_{AgCl}=0,25.143,5=35,875\left(g\right)\)
c. Theo PT: \(n_{HCl}=n_{AgCl}=0,25\left(mol\right)\)
Đổi 100ml = 0,1 lít
=> \(C_{M_{HCl}}=\dfrac{0,25}{0,1}=2,5M\)
a) \(n_{Fe}=\dfrac{1,12}{56}=0,02\left(mol\right)\)
PTHH: Fe + 2HCl -->FeCl2 + H2
_____0,02->0,04--->0,02--->0,02
=> VH2 = 0,02.22,4 = 0,448(l)
b) mFeCl2 = 0,02.127 = 2,54(g)
c) \(C_{M\left(HCl\right)}=\dfrac{0,04}{0,2}=0,2M\)
Fe + 2HCl → FeCl2 + H2
1 2 1 1
0,02 0,04 0,02 0,02
nFe=\(\dfrac{1,12}{56}\)= 0,02(mol)
a). nH2=\(\dfrac{0,02.1}{1}\)= 0,02(mol)
→VH2= n . 22,4 = 0,02 . 22,4 = 0,448(l)
b). nFeCl2= \(\dfrac{0,02.1}{1}\)= 0,02(mol)
→mFeCl2= n . M = 0,02 . 127 = 2,54(g)
c). 200ml = 0,2l
nHCl= \(\dfrac{0,02.2}{1}\)=0,04(mol)
→CM= \(\dfrac{n}{V}\)= \(\dfrac{0,04}{0,2}\)= 0,2M
\(n_{Na_2SO_3}=\dfrac{12,6}{126}=0,1mol\)
\(Na_2SO_3+H_2SO_4\rightarrow Na_2SO_4+SO_2+H_2O\)
a) \(n_{SO_2}=n_{Na_2SO_3}=0,1mol\) \(\Rightarrow V=2,24l\)
b) \(n_{H_2SO_4}=n_{Na_2SO_3}=0,1mol\) \(\Rightarrow C_M=\dfrac{0,1}{0,2}=0,5M\)
c) \(m_{Na_2SO_4}=0,1\cdot142=14,2g\)
a)
$n_{Na_2O} = \dfrac{15,5}{62} = 0,25(mol)$
$Na_2O + H_2O \to 2NaOH$
$n_{NaOH} = 2n_{Na_2O} = 0,5(mol)$
$C_{M_{NaOH}} = \dfrac{0,5}{0,5} = 1M$
b)
$2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$
$n_{H_2SO_4} = \dfrac{1}{2}n_{NaOH} = 0,25(mol)$
$\Rightarrow m_{dd\ H_2SO_4} = \dfrac{0,25.98}{20\%} = 122,5(gam)$
$\Rightarrow V_{dd\ H_2SO_4} = \dfrac{122,5}{1,14} = 107,46(ml)$
\(n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
PT: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)
Theo PT: \(n_{Ca\left(OH\right)_2}=n_{CaCO_3}=n_{CO_2}=0,25\left(mol\right)\)
a, \(C_{M_{Ca\left(OH\right)_2}}=\dfrac{0,25}{0,1}=2,5\left(M\right)\)
b, \(m_{CaCO_3}=0,25.100=25\left(g\right)\)
c, \(Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\)
Theo PT: \(n_{HCl}=2n_{Ca\left(OH\right)_2}=0,5\left(mol\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{0,5.36,5}{20\%}=91,25\left(g\right)\)
\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ n_{AlCl_3}=n_{Al}=0,1\left(mol\right)\\ a,m_{AlCl_3}=133,5.0,1=13,35\left(g\right)\\ n_{H_2}=\dfrac{3}{2}.0,1=0,15\left(mol\right)\\ b,V_{H_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\\ c,n_{HCl}=\dfrac{6}{2}.0,1=0,3\left(mol\right)\\ c,C_{MddHCl}=\dfrac{0,3}{0,2}=1,5\left(M\right)\)
PTHH\(Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\)
tl............1................2.............2.............1.............1..(mol
br 0,1.................0,2......................................0,1(mol)
NaCl không phản ứng đc vsHCl
b)\(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(C_{MHCl}=\dfrac{0,2}{0,4}=0,5\left(M\right)\)(đổi 400ml=0,4(l))
c)\(Tacom_{Na_2CO_3}=0,1.106=10,6\left(g\right)\)
\(\Rightarrow\%m_{Na_2CO_3}=\dfrac{10,6}{20}.100=53\%\)
\(\Rightarrow\%mNaCl=100\%-53\%=47\%\)
\(n_{Na2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\)
Pt : \(Na_2O+2HCl\rightarrow2NaCl+H_2O|\)
1 2 2 1
0,25 0,5 0,5
a) \(n_{NaCl}=\dfrac{0,25.2}{1}=0,5\left(mol\right)\)
⇒ \(m_{NaCl}=0,5.58,5=29,25\left(g\right)\)
b) \(n_{HCl}=\dfrac{0,25.2}{1}=0,5\left(mol\right)\)
\(C_{M_{ddHCl}}=\dfrac{0,5}{0,5}=1\left(M\right)\)
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0,5 lit đ HCl em nhầm tí ạ