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\(m_{H_2SO_4}=\dfrac{200\cdot9,8\%}{100\%}=19,6\left(g\right)\\ n_{H_2SO_4}=\dfrac{19,6}{98}=0,2\left(mol\right)\\ PTHH:2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\\ \Rightarrow n_{KOH}=2n_{H_2SO_4}=0,4\left(mol\right)\\ \Rightarrow m_{KOH}=0,4\cdot\left(39+16+1\right)=22,4\left(g\right)\)
\(m_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\
m_{H_2SO_4}=200.12,25\%=24,5g\\
n_{H_2SO_4}=\dfrac{24,5}{98}=0,25\\
pthh:Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\\
LTL:0,2< 0,25=>H_2SO_4\left(d\text{ư}\right)Fe\left(h\text{ết}\right)\)
\(n_{H_2}=n_{Fe}=0,2\left(mol\right)\\
V_{H_2}=0,2.22,4=4,48l\)
\(C\%=\dfrac{11,2}{24,5}.100\%=45,7\%\)
a)
Gọi $n_{NaOH} = a(mol) ; n_{KOH} = b(mol) \Rightarrow 40a + 56b = 3,04(1)$
$NaOH + HCl \to NaCl + H_2O$
$KOH + HCl \to KCl + H_2O$
$m_{muối} = 58,5a + 74,5b = 4,15(2)$
Từ (1)(2) suy ra a = 0,02 ; b = 0,04
$n_{HCl} = a + b = 0,06(mol)$
$C\%_{HCl} = \dfrac{0,06.36,5}{200}.100\% = 1,095\%$
b)
$m_{dd} = 3,04 + 200 = 203,4(gam)$
$C\%_{NaCl} = \dfrac{0,02.58,5}{203,4}.100\% = 0,58\%$
$C\%_{KCl} =\dfrac{0,04.74,5}{203,4}.100\% = 1,47\%$
\(m_{HCl}=\dfrac{300.7,3\%}{100\%}=21,9g\\ n_{HCl}=\dfrac{21,9}{36,5}=0,6mol\\ HCl+NaOH\rightarrow NaCl+H_2O\left(1\right)\\ n_{NaOH\left(1\right)}=n_{HCl}=0,6mol\\ m_{H_2SO_4}=\dfrac{200.9,8\%}{100\%}=19,6g\\ n_{H_2SO_4}=\dfrac{19,6}{98}=0,2mol\\ H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\left(2\right)\\ n_{NaOH\left(2\right)}=0,2.2=0,4mol\\ n_{NaOH}=0,4+0,6=1mol\\ m_{NaOH}=1.40=40g\\ m_{ddNaOH}=\dfrac{40}{5\%}\cdot100\%=800g\)
mH2SO4=200.9,8%=19,6(g) -> nH2SO4=19,6/98=0,2(mol)
mKOH=200.5,6%=11,2(g) -> nKOH=11,2/56=0,2(mol)
PTHH: 2 KOH + H2SO4 -> K2SO4 + 2 H2O
Ta có: 0,2/2 < 0,2/1
=> H2SO4 dư, KOH hết, tính theo nKOH
=> Chất có trong dd thu được sau p.ứ: K2SO4 và H2SO4(dư)
nH2SO4(p.ứ)=nK2SO4=1/2. nKOH=1/2. 0,2=0,1(mol)
nH2SO4(dư)=0,2-0,1=0,1(mol) => mH2SO4(dư)=0,1.98=9,8(g)
mK2SO4=174. 0,1=17,4(g)
mddsau= mddH2SO4 + mddKOH= 200+200=400(g)
=>C%ddH2SO4(dư)= (9,8/400).100=2,45%
C%ddK2SO4=(17,4/400).100=4,35%
\(m_{H_2SO_4}=19,6\left(g\right)\Rightarrow n_{H_2SO_4}=0,2\left(mol\right)\)
\(m_{KOH}=11,2\left(g\right)\Rightarrow n_{KOH}=0,2\left(mol\right)\)
PTHH: \(H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O\)
\(\Rightarrow H_2SO_4\) dư.
\(\Rightarrow n_{H_2SO_4dư}=0,3\left(mol\right);n_{K_2SO_4}=0,1\left(mol\right)\)
\(\Rightarrow C\%_{H_2SO_4dư}=\dfrac{0,3.98}{400}.100\%=7,35\%\)
\(C\%_{K_2SO_4}=\dfrac{0,1.174}{400}.100\%=4,35\%\)
Bài 1:
Ta có: \(n_{HCl}=0,2.0,5=0,1\left(mol\right)\)
BTNT H, có: \(n_{HCl}=2n_{H_2O}\Rightarrow n_{H_2O}=0,05\left(mol\right)\)
Theo ĐL BTKL, có: m oxit + mHCl = mmuối + mH2O
⇒ mmuối = 2,8 + 0,1.36,5 - 0,05.18 = 5,55 (g)
Bài 2:
\(m_{KOH}=200.5,6\%=11,2\left(g\right)\Rightarrow n_{KOH}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
PT: \(2KOH+CuCl_2\rightarrow2KCl+Cu\left(OH\right)_2\)
Theo PT: \(n_{Cu\left(OH\right)_2}=\dfrac{1}{2}n_{KOH}=0,1\left(mol\right)\)
\(\Rightarrow m_{Cu\left(OH\right)_2}=0,1.98=9,8\left(g\right)\)