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a)
$Mg + 2HCl \to MgCl_2 + H_2$
$MgO + 2HCl \to MgCl_2 + H_2o$
b)
Theo PTHH : $n_{Mg} = n_{H_2} = \dfrac{4,48}{22,4} = 0,2(mol)$
$m_{Mg} = 0,2.24 = 4,8(gam)$
$m_{MgO} = m_{hh} - m_{Mg} = 12,8 - 4,8 = 8(gam)$
c)
$n_{MgO} = \dfrac{8}{40} = 0,2(mol)$
$n_{HCl} = 2n_{Mg} + 2n_{MgO} = 0,8(mol)$
$m_{dd\ HCl} = \dfrac{0,8.36,5}{14,6\%} = 200(gam)$
PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(MgO+2HCl\rightarrow MgCl_2+H_2O\)
a, Ta có: \(n_{H_2}=0,1\left(mol\right)\)
Theo PT: \(n_{Mg}=n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow m_{Mg}=0,1.24=2,4\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{2,4}{4,4}.100\%\approx54,55\%\\\%m_{MgO}\approx45,45\%\end{matrix}\right.\)
b, Ta có: mMgO = mhhA - mMg = 2 (g)
\(\Rightarrow n_{MgO}=\dfrac{2}{40}=0,05\left(mol\right)\)
Theo PT: \(n_{HCl}=2n_{MgO}=0,1\left(mol\right)\)
\(\Rightarrow V_{HCl}=\dfrac{0,1}{2}=0,05\left(l\right)=50\left(ml\right)\)
Bạn tham khảo nhé!
nH2=0,1(mol)
PTHH: Mg + 2 HCl -> MgCl2 + H2
0,1__________0,2___________0,1(mol)
MgO + 2 HCl -> MgCl2 + H2O
0,05____0,1___0,05(mol)
mMg=0,1. 24= 2,4(g) -> mMgO=4,4-2,4= 2(g) -> nMgO=0,05((mol)
b) %mMg= (2,4/4,4).100=54,545%
=> %mMgO=45,455%
c) nHCl=0,3(mol) -> mHCl=0,3.36,5=10,95(g)
=> mddHCl=(10,95.100)/7,3=150(g)
Câu 4 :
\(n_{H2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
a) Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)
1 2 1 1
0,05 0,1 0,05
\(MgO+2HCl\rightarrow MgCl_2+H_2O|\)
1 2 1 1
0,2 0,4
b) \(n_{Mg}=\dfrac{0,05.1}{1}=0,05\left(mol\right)\)
\(m_{Mg}=0,05.24=1.2\left(g\right)\)
\(m_{MgO}=9,2-1,2=8\left(g\right)\)
c) Có : \(m_{MgO}=8\left(g\right)\)
\(n_{MgO}=\dfrac{8}{40}=0,2\left(mol\right)\)
\(n_{HCl\left(tổng\right)}=0,1+0,4=0,5\left(mol\right)\)
\(m_{HCl}=0,05.36,5=18,25\left(g\right)\)
\(m_{ddHCl}=\dfrac{18,25.100}{14,6}=125\left(g\right)\)
Chúc bạn học tốt
Bạn ơi cho mik hỏi, tại sao nH2 lại là o,o5 mol v ? 1,12/22,4 là bằng 0,1 ....vậy tại sao lại ra 0,05 v ?
\(n_{CO2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PTHH : \(CaO+2HCl\rightarrow CaCl_2+H_2O\) (1)
\(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\) (2)
a) Theo Pt : \(n_{CO2}=n_{CaCO3}=0,1\left(mol\right)\)
\(m_{CaCO3}=0,1100=10\left(g\right)\)
\(m_{CaO}=12,8-10=2,8\left(g\right)\)
b) Chắc tính V của dd HCl đã dùng
(1) \(n_{CaO}=\dfrac{2,8}{56}=0,05\left(mol\right)\) , \(n_{HCl}=2n_{CaO}=0,1\left(mol\right)\)
(2) \(n_{HCl}=2n_{CaCO3}=0,2\left(mol\right)\)
\(\Rightarrow V_{ddHCl}=\dfrac{0,1+0,2}{1}=0,3\left(l\right)=300\left(ml\right)\)
a, \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
\(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)
b, \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Theo PT: \(n_{Mg}=n_{H_2}=0,1\left(mol\right)\Rightarrow m_{Mg}=0,1.24=2,4\left(g\right)\)
\(\Rightarrow m_{MgO}=4,4-2,4=2\left(g\right)\)
c, \(n_{MgO}=\dfrac{2}{40}=0,05\left(mol\right)\)
Theo PT: \(n_{H_2SO_4}=n_{Mg}+n_{MgO}=0,15\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,15.98=14,7\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{14,7}{19,6\%}=75\left(g\right)\)
Đặt \(n_{Mg}=x;n_{MgO}=y\)
\(PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\)
(mol) 1 2 1 1
(mol) x 2x x x
\(PTHH:MgO+2HCl\rightarrow MgCl_2+H_2O\)
(mol) 1 2 1 1
(mol) y 2y y y
\(hpt:\left\{{}\begin{matrix}24x+40y=4,4\\22,4x=2,24\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}n_{Mg}=0,1\left(mol\right)\rightarrow m_{Mg}=0,1.24=2,4\left(g\right)\rightarrow\%m_{Mg}=\frac{2,4}{4,4}.100\%=54,54\left(\%\right)\\n_{MgO}=0,05\left(mol\right)\rightarrow m_{MgO}=0,05.40=2\left(g\right)\rightarrow\%m_{MgO}=100\%-54,54\%=45,46\left(\%\right)\end{matrix}\right.\)
\(m_{ddHCl}=\frac{36,5.\left(2.0,1+2.0,05\right).100\%}{14,6\%}=75\left(g\right)\)
\(m_{H_2}=x.2=0,1.2=0,2\left(g\right)\)
\(C\%_{ddspu}=\frac{\left(0,1+0,05\right).95}{4,4+75-0,2}.100\%=18\left(\%\right)\)
Mg + 2HCl → MgCl2 + H2 (1)
MgO + 2HCl → MgCl2 + H2O (2)
\(n_{H_2}=\frac{2,24}{22,4}=0,1\left(mol\right)\)
a) Theo Pt1: \(n_{Mg}=n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow m_{Mg}=0,1\times24=2,4\left(g\right)\)
\(\Rightarrow m_{MgO}=4,4-2,4=2\left(g\right)\)
\(\%m_{Mg}=\frac{2,4}{4,4}\times100\%=54,55\%\)
\(\%m_{MgO}=\frac{2}{4,4}\times100\%=45,45\%\)
b) \(m_{H_2}=0,1\times2=0,2\left(g\right)\)
Theo pT1: \(n_{HCl}=2n_{Mg}=2\times0,1=0,2\left(mol\right)\)
\(n_{MgO}=\frac{2}{40}=0,05\left(mol\right)\)
Theo Pt2: \(n_{HCl}=2n_{MgO}=2\times0,05=0,1\left(mol\right)\)
\(\Rightarrow\Sigma n_{HCl}=0,2+0,1=0,3\left(mol\right)\)
\(\Rightarrow\Sigma m_{HCl}=0,3\times36,5=10,95\left(g\right)\)
\(\Rightarrow m_{ddHCl}=\frac{10,95}{14,6\%}=75\left(g\right)\)
Ta có: \(m_{dd}saupứ=4,4+75-0,2=79,2\left(g\right)\)
Theo pt1: \(n_{MgCl_2}=n_{Mg}=0,1\left(mol\right)\)
Theo Pt2: \(n_{MgCl_2}=n_{MgO}=0,05\left(mol\right)\)
\(\Rightarrow\Sigma n_{MgCl_2}=0,1+0,05=0,15\left(mol\right)\)
\(\Rightarrow\Sigma m_{MgCl_2}=0,15\times95=14,25\left(g\right)\)
\(\Rightarrow C\%_{MgCl_2}=\frac{14,25}{79,2}\times100\%=17,99\%\)