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a) $Ba(OH)_2 + 2HCl \to BaCl_2 + 2H_2O$
$n_{HCl} = 0,2.0,5 = 0,1(mol)$
$n_{BaCl_2} = \dfrac{1}{2}n_{HCl} = 0,05(mol)$
$m_{BaCl_2} = 0,05.208 = 10,4(gam)$
b) $n_{Ba(OH)_2} = n_{BaCl_2} = 0,05(mol)$
$\Rightarrow V = \dfrac{0,05}{2} = 0,025(lít) = 25(ml)$
Đổi 200ml = 0,2 lít
Ta có: \(n_{HCl}=0,5.0,2=0,1\left(mol\right)\)
a. PTHH: 2HCl + Ba(OH)2 ---> BaCl2 + 2H2O
Theo PT: \(n_{BaCl_2}=\dfrac{1}{2}.n_{HCl}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\)
=> \(m_{BaCl_2}=0,05.208=10,4\left(g\right)\)
b. Theo PT: \(n_{Ba\left(OH\right)_2}=n_{BaCl_2}=0,05\left(mol\right)\)
=> \(V_{dd_{Ba\left(OH\right)_2}}=\dfrac{0,05}{2}=0,025\left(lít\right)=25\left(ml\right)\)
\(n_{H_2SO_4}=1.0,2=0,2(mol)\\ PTHH:2NaOH+H_2SO_4\to Na_2SO_4+2H_2O\\ a,n_{NaOH}=0,4(mol);n_{Na_2SO_4}=0,2(mol)\\ \Rightarrow \begin{cases} m_{Na_2SO_4}=0,2.142=28,4(g)\\ m_{dd_{NaOH}}=\dfrac{0,4.40}{20\%}=80(g) \end{cases}\\ b,2KOH+H_2SO_4\to K_2SO_4+2H_2O\\ \Rightarrow n_{KOH}=0,4(mol)\\ \Rightarrow m_{dd_{KOH}}=\dfrac{0,4.56}{5,6\%}=400(g)\\ \Rightarrow V_{dd_{KOH}}=\dfrac{400}{1,045}=382,78(ml)\)
Bước 1: nH2SO4 = VH2SO4 . CM H2SO4= 0,2 . 1 = 0,2mol
Bước 2:
PTHH: 2NaOH + H2SO4 → Na2SO4 + H2O
2 mol 1 mol
? mol 0,2mol
nNaOH=0,2.21=0,4mol.nNaOH=0,2.21=0,4mol.
m NaOH= n NaOH.MNaOH = 0,4 . (23 + 16 + 1) = 16g
Bước 3: C% = mNaOH : m dd NaOH => mdd NaOH = mNaOH : C% = 16 : 20% = 80g
a, Ta có :
nBa(OH)2 : nNaOH = 1:2
Gọi số mol của Ba(OH)2 là x mol nên số mol của NaOH là 2x mol
nHCl=0,4(mol)
PTHH: NaOH + HCl -> NaCl +H2O
_____2x_________________________
PTHH: Ba(OH)2 + 2HCl -> BaCl2 + 2H2O
________x____________________
nHCl = nNaOh + 2nBa(OH)2 = 2x + 2x = 0,4mol
=> x = 0,1 (mol)
=> nNaOH = 0,1.2 = 0,2 mol
=> CM NaOH = 0,2/0,2 = 1M
=> nBa(OH)2 = 0,1 (mol) => CMBa(OH)2 = 0,1/0,2 = 0,5M
b, nNaCl = nNaOH = 0,2mol
=> nBaCl2 = nBa(OH)2 = 0,1mol
=> m muối = 0,2.58,5 + 0,2.208 = 53,24g
200ml = 0,2l
\(n_{Ba\left(OH\right)2}=0,5.0,2=0,1\left(mol\right)\)
Pt : \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O|\)
1 2 1 2
0,1 0,2 0,1
a) \(n_{HCl}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)
\(V_{ddHCl}=\dfrac{0,2}{1}=0,2\left(l\right)=200\left(ml\right)\)
b) \(n_{BaCl2}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)
⇒ \(m_{BaCl2}=0,1.208=20,8\left(g\right)\)
c) \(V_{ddspu}=0,2+0,2=0,4\left(l\right)\)
\(C_{M_{BaCl2}}=\dfrac{0,1}{0,4}=0,25\left(M\right)\)
Chúc bạn học tốt
PTHH: \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)
Ta có: \(n_{Ba\left(OH\right)_2}=0,2\cdot0,5=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,2\left(mol\right)\\n_{BaCl_2}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{ddHCl}=\dfrac{0,2}{1}=0,2\left(l\right)=200\left(ml\right)\\m_{BaCl_2}=0,1\cdot208=20,8\left(g\right)\\C_{M_{BaCl_2}}=\dfrac{0,1}{0,2+0,2}=0,25\left(M\right)\end{matrix}\right.\)
a)
\(n_{NaOH}\)=0,2.5=1(mol)
\(n_{Ba\left(OH\right)_2}\)=0,2.2=0,4(mol)
\(NaOH+HCL\rightarrow NaCl+H_2O\)(1)
1 1 1 1 (mol)
\(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\left(2\right)\)
0,4 0,8 0,4 0,8 (mol)
\(m_{NaCl}\)=1.58,5=58,5g
\(m_{BaCl_2}\)=0,4.208=83,2g
b)
\(m_{HCl}\)=36,5.(1+0,8)=65,7g
\(m_{ddHCl}\)=\(\frac{65,7.100}{12.4}\)=530g
\(V_{ddHCl}=\frac{530}{1,06}\)=500ml