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a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)
b, \(n_{Mg}=\dfrac{1,2}{24}=0,05\left(mol\right)\)
Theo PT: \(n_{HCl}=2n_{Mg}=0,1\left(mol\right)\Rightarrow m_{ddHCl}=\dfrac{0,1.36,5}{10\%}=36,5\left(g\right)\)
c, \(n_{H_2}=n_{MgCl_2}=n_{Mg}=0,05\left(mol\right)\)
Ta có: m dd sau pư = 1,2 + 36,5 - 0,05.2 = 37,6 (g)
\(\Rightarrow C\%_{MgCl_2}=\dfrac{0,05.95}{37,6}.100\%\approx12,63\%\)
a)
\(m_{H_2SO_4}=\dfrac{300.19,6}{100}=58,8\left(g\right)\)
=> \(m_{dd.H_2SO_4.9,8\%}=\dfrac{58,8.100}{9,8}=600\left(g\right)\)
=> \(m_{H_2O\left(thêm\right)}=600-300=300\left(g\right)\)
b)
\(n_{HCl}=0,2.2=0,4\left(mol\right)\)
=> \(V_{dd.HCl.1,5M}=\dfrac{0,4}{1,5}=\dfrac{4}{15}\left(l\right)\)
=> \(V_{H_2O\left(thêm\right)}=\dfrac{4}{15}-0,2=\dfrac{1}{15}\left(l\right)=\dfrac{200}{3}\left(ml\right)\)
=> \(m_{H_2O\left(thêm\right)}=\dfrac{200}{3}.1=\dfrac{200}{3}\left(g\right)\)
\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
\(0.2.......0.2......................0.2\)
\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)
\(C\%H_2SO_4=\dfrac{0.2\cdot98\cdot100\%}{200}=9.8\%\)
\(n_{Mg}=\dfrac{2.4}{24}=0.1\left(mol\right)\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(0.1.......0.2...........0.1........0.1\)
\(m_{dd_{HCl}}=\dfrac{0.2\cdot36.5\cdot100}{14.6}=50\left(g\right)\)
\(V_{H_2}=0.1\cdot22.4=2.24\left(l\right)\)
\(m_{MgCl_2}=0.1\cdot95=9.5\left(g\right)\)
\(m_{\text{dung dịch sau phản ứng}}=2.4+50-0.1\cdot2=52.2\left(g\right)\)
\(C\%_{MgCl_2}=\dfrac{9.5}{52.2}\cdot100\%=18.2\%\)
\(a.\\ C\%_{sau}=\dfrac{5}{100}=\dfrac{32.0,1}{32+m_{H_2O}}\\ m_{H_2O}=32\left(g\right)\\ b.\\ C_{M\left(sau\right)}=1=\dfrac{0,2.2}{0,2+V_{H_2O}}\\ V_{H_2O}=0,2\left(L\right)=200\left(mL\right)\)
nFe = 2,8 : 56 = 0,05 ( mol )
PTHH : Fe + 2HCl -----> FeCl2 + H2
mol 0,05 0,1 0,05
VH2 = 0,05 x 22,4 = 1,12 ( l )
mHCl = 0,1 x 36,5 = 3,65 ( g)
=> mHCl (10%) = 3,65 x 100 : 10 = 36,5 (g)
PTHH
Fe + 2HCl \(\rightarrow\) Fe + H2O
gt 0,05 0,1 0,05 0,05
mFe = 2,8 g \(\Rightarrow\) nFe = \(\frac{m}{M}\)= \(\frac{2,8}{56}=0,05\left(mol\right)\)
Theo ptpư + gt ta có:
V\(H_2\) = n. 22,4 = 0,05 . 22,4 = 1,12 (lít)
mHCl = 0,1 . 36,5 = 3, 65 (g)
mdd HCl = \(\frac{3,65.100}{10}=36,5\left(g\right)\)
a) Fe + 2HCl --> FeCl2 + H2
b) \(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
0,05<-0,1<-----------0,05
=> m = 0,05.56 = 2,8 (g)
c) \(m_{HCl}=0,1.36,5=3,65\left(g\right)\Rightarrow m_{dd.HCl}=\dfrac{3,65.100}{10}=36,5\left(g\right)\)
a) n Fe = 28/56 = 0,5(mol)
$Fe + 2HCl \to FeCl_2 + H_2$
Theo PTHH :
n HCl = 2n Fe = 1(mol)
=> m dd HCl = 1.36,5/10% = 365(gam)
b)
n FeCl2 = n H2 = n Fe = 0,5(mol)
Suy ra :
V H2 = 0,5.22,4 = 11,2(lít)
m FeCl2 = 0,5.127 = 63,5(gam)
c)
Sau phản ứng:
mdd = m Fe + mdd HCl - m H2 = 28 + 365 - 0,5.2 = 392(gam)
=> C% FeCl2 = 63,5/392 .100% = 16,2%
\(m_{HCl}=\dfrac{600.10\%}{100\%}=60\left(g\right)\)
Gọi `x` là m gam HCl cần thêm vào, có:
\(C\%_{HCl}=\dfrac{\left(x+60\right).100\%}{600}=18\%\\ \Rightarrow x=48\)
Vậy m HCl cần thêm vào 600g dung dịch HCl `10%` để được dung dịch HCl `18%` là 48 (g)
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