Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)
b, \(m_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\)
Theo PT: \(n_{MgCl_2}=n_{Mg}=0,3\left(mol\right)\Rightarrow m_{MgCl_2}=0,3.95=28,5\left(g\right)\)
c, \(n_{HCl}=2n_{Mg}=0,6\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,6}{0,2}=3\left(M\right)\)
\(a)\\ Fe + 2HCl \to FeCl_2 + H_2\)
b)
\(n_{Fe} = \dfrac{22,4}{56}= 0,4(mol)\\ n_{H_2} = \dfrac{6,72}{22,4} = 0,3(mol)\)
Ta thấy : \(n_{Fe} > n_{H_2}\) nên Fe dư.
Theo PTHH :
\(n_{Fe\ pư} = n_{H_2} = 0,3(mol)\\ \Rightarrow m_{Fe\ pư} = 0,3.56 = 16,8(gam)\)
c)
Ta có :
\(n_{FeCl_2} = n_{H_2} = 0,3(mol)\\ \Rightarrow m_{FeCl_2} = 0,3.127 = 38,1(gam)\)
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)PTHH: Zn + 2HCl -> ZnCl_2 + H_2↑`
____________________________________
`b) n_[Zn] = [ 6,5 ] / 65 = 0,1 (mol)`
Theo `PTHH` có: `n_[H_2] = n_[Zn] = 0,1 (mol)`
`-> V_[H_2 (đktc)] = 0,1 . 22,4 = 2,24 (l)`
______________________________________
`c)` Theo `PTHH` có: `n_[HCl] = 2 n_[Zn] = 2 . 0,1 = 0,2 (mol)`
Đổi `200 ml = 0,2 l`
`-> C_[M_[HCl]] = [ 0,2 ] / [0,2 ] = 1(M)`
________________________________________
`d)` Theo `PTHH` có: `n_[ZnCl_2] = n_[Zn] = 0,1 (mol)`
`-> m_[ZnCl_2] = 0,1 . 136 = 13,6 (g)`
VHCl= 200ml = 0,2 (l)
Zn + 2HCl -- > ZnCl2 + H2
nZn = 6,5 : 65 = 0,1(mol)
VH2 = 0,1 . 22,4 = 2,24 (l)
\(C_{MHCl}=\dfrac{n}{V}=\dfrac{0,2}{0,2}=1M\)
mZnCl2 = 0,1 . 136 = 13,6 (g)
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
PTHH :
\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
0,2 0,4 0,2 0,2
\(a,V_{H_2}=0,2.22,4=4,48\left(l\right)\)
\(b,m_{HCl}=0,4.36,5=14,6\left(g\right)\)
\(m_{ddHCl}=\dfrac{14,6.100}{10}=146\left(g\right)\)
\(c,m_{MgCl_2}=0,2.95=19\left(g\right)\)
\(m_{ddMgCl_2}=4,8+146-\left(0,2.2\right)=150,4\left(g\right)\)
\(C\%_{MgCl_2}=\dfrac{19}{150,4}.100\%\approx12,63\%\)
2.
\(n_K=\dfrac{7,8}{39}=0,2\left(mol\right)\)
\(2K+2H_2O\rightarrow2KOH+H_2\uparrow\)
0,2 0,2 0,1
\(m_{KOH}=0,2.56=11,2\left(g\right)\)
\(m_{ddKOH}=7,8+100-\left(0,1.2\right)=107,6\left(g\right)\)
\(C\%=\dfrac{11,2}{107,6}.100\%\approx10,4\%\)
Zn+2HCl->ZnCl2+H2
0,2-----0,4---0,2----0,2
nZn=0,2 mol
=>m Hcl=0,4.36,5=14,6g
m muối=0,2.136=27,2g
=>VH2=0,2.22,4=4,48l
`Zn + 2HCl -> ZnCl_2 + H_2` `\uparrow`
`n_(Zn) = 13/65 = 0,2 mol`.
`n_(HCl) = 0,4 mol`.
`m_(HCl) = 0,4 xx 36,5 = 14,6g`.
c, `m_(ZnCl_2) = 0,2 xx 127 = 25,4 g`.
`d, V_(H_2) = 0,2 xx 22,4 = 4,48l`.
PTHH :
\(Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\uparrow\)
x 2x 2x x x
\(K_2CO_3+2HCl\rightarrow2KCl+H_2O+CO_2\uparrow\)
y 2y 2y y y
Ta có :
106x + 138y = 26
2x + 2y = 0,4
Giải hệ PT, ta có :
\(\rightarrow x=0,05\left(mol\right);y=0,15\left(mol\right)\)
Thu đc khí CO2 chứ bạn nhỉ?
\(a,V_{CO_2}=\left(0,05+0,15\right).22,4=4,48\left(l\right)\)
\(b,m_{muối}=0,05.58,5+0,15.74,5=14,1\left(g\right)\)
\(c,\%m_{Na_2CO_3}=\dfrac{0,05.106}{26}.100\%\approx20,38\%\)
\(\%m_{K_2CO_3}=100\%-20,38\%=79,62\%\)
a) Zn + 2HCl --> ZnCl2 + H2
b) \(n_{H_2}=\dfrac{6,1975}{24,79}=0,25\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
0,25<-0,5<---0,25<--0,25
=> mZn = 0,25.65 = 16,25 (g)
c) mZnCl2 = 0,25.136 = 34 (g)
\(a)Mg+2HCl\rightarrow MgCl_2+H_2\\ b)n_{HCl}=0,2.1=0,2mol\\ m_{HCl}=0,2.36,5=7,3g\\ c)n_{H_2}=\dfrac{2,479}{24,79}=0,1mol\\ BTKL:m_{Mg}+m_{HCl}=m_{MgCl_2}+m_{H_2}\\ \Rightarrow m_{MgCl_2}=2,4+7,3-0,1.2=9,5g\)