Bài 1
Hòa tan hết 4,4g h2 Mg,MgO cần 300ml dung dịch HCL được 2,24 lít khí (đktc)
a)%khối lượng từng chất trong hỗn hợp
b)Cm axit
c)Cm dung dịch sau phản ứng
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Khách
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Những câu hỏi liên quan
30 tháng 9 2018
Mg + 2HCl → MgCl2 + H2 (1)
MgO + 2HCl → MgCl2 + H2O (2)
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Theo PT1: \(n_{Mg}=n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow m_{Mg}=0,1\times24=2,4\left(g\right)\)
\(\Rightarrow m_{MgO}=4,4-2,4=2\left(g\right)\)
\(\Rightarrow\%Mg=\dfrac{2,4}{4,4}\times100\%=54,55\%\)
\(\%MgO=\dfrac{2}{4,4}\times100\%=45,45\%\)
Theo PT1: \(n_{HCl}=2n_{Mg}=2\times0,1=0,2\left(mol\right)\)
\(n_{MgO}=\dfrac{2}{40}=0,05\left(mol\right)\)
Theo PT2: \(n_{HCl}=2n_{MgO}=2\times0,05=0,1\left(mol\right)\)
\(\Rightarrow\Sigma n_{HCl}=0,2+0,1=0,3\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,3}{0,3}=1\left(M\right)\)
Theo PT1,2: \(\Sigma n_{MgCl_2}=\dfrac{1}{2}\Sigma n_{HCl}=\dfrac{1}{2}\times0,3=0,15\left(mol\right)\)
\(\Rightarrow C_{M_{MgCl_2}}=\dfrac{0,15}{0,3}=0,5\left(M\right)\)
26 tháng 7 2021
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)
H
25 tháng 12 2023
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
a.
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
b.
\(n_{Mg}=n_{H_2}=0,1\left(mol\right)\Rightarrow m_{Mg}=0,1.24=2,4\left(g\right),m_{CuO}=12-2,4=9,6\left(g\right)\)
c.
\(m_{muối}=m_{CuCl_2}+m_{MgCl_2}=0,12.135+95.0,1=25,7\left(g\right)\)
27 tháng 12 2020
PT: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\left(1\right)\)
\(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\left(2\right)\)
Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Theo PT (1): \(n_{Mg}=n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow m_{Mg}=0,1.24=2,4\left(g\right)\)
\(\Rightarrow m_{MgO}=m_{hh}-m_{Mg}=4,4-2,4=2\left(g\right)\)
Bạn tham khảo nhé!
21 tháng 12 2020
PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
\(MgO+2HCl\rightarrow MgCl_2+H_2O\)
Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)=n_{Mg}\)
\(\Rightarrow m_{Mg}=0,1\cdot24=2,4\left(g\right)\) \(\Rightarrow m_{MgO}=2\left(g\right)\)
M
13 tháng 10 2021
a,
Mg+ 2HCl= MgCl2+ H2
MgO+ 2HCl= MgCl2+ H2O
b,
nH2= 2,24/22,4= 0,1 mol
=> nMg= nMgCl2= 0,5nHCl= 0,1 mol => nHCl= 0,2 mol
=> mMg= 0,1.24= 2,4g
=> mMgO= 2g
c,
nMgO= 2/40= 0,05 mol
=> nMgO= 0,5nHCl= nMgCl2= 0,05 mol
=> nHCl= 0,1 mol
Tổng lượng HCl cần dùng là 0,1+0,2=0,3 mol
=> m dd HCl= 0,3.36,5.100:7,3= 150g
27 tháng 10 2021
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
0,1 0,2 0,1 (mol)
\(\Rightarrow m_{Mg}=0,1\cdot24=2,4\left(g\right)\)
\(\Rightarrow m_{MgO}=4,4-2,4=2\left(g\right)\)\(\Rightarrow n_{MgO}=\dfrac{2}{40}=0,05mol\)
\(MgO+2HCl\rightarrow MgCl_2+H_2O\)
0,05 0,1
\(\Rightarrow\Sigma n_{HCl}=0,2+0,1=0,3mol\)\(\Rightarrow m_{HCl}=0,3\cdot36,5=10,95\left(g\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{10,95}{7,3}\cdot100=150\left(g\right)\)
HP
1 tháng 11 2021
Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
a. PTHH:
Mg + 2HCl ---> MgCl2 + H2 (1)
MgO + 2HCl ---> MgCl2 + H2O (2)
Theo PT(1): \(n_{Mg}=n_{H_2}=0,1\left(mol\right)\)
=> \(m_{Mg}=0,1.24=2,4\left(g\right)\)
=> \(m_{MgO}=4,4-2,4=2\left(g\right)\)
b. Ta có: \(n_{MgO}=\dfrac{2}{40}=0,05\left(mol\right)\)
=> \(n_{hh}=0,05+0,1=0,15\left(mol\right)\)
Theo PT(1,2): \(n_{HCl}=2.n_{hh}=2.0,15=0,3\left(mol\right)\)
=> \(V_{dd_{HCl}}=\dfrac{0,3}{0,4}=0,75\left(lít\right)\)
Mg + 2HCl → MgCl2 + H2 (1)
MgO + 2HCl → MgCl2 + H2O (2)
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Theo PT1: \(n_{Mg}=n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow m_{Mg}=0,1\times24=2,4\left(mol\right)\)
\(\Rightarrow m_{MgO}=4,4-2,4=2\left(g\right)\)
\(\Rightarrow\%Mg=\dfrac{2,4}{4,4}\times100\%=54,55\%\)
\(\%MgO=\dfrac{2}{4,4}\times100\%=45,45\%\)
Theo PT1: \(n_{HCl}=2n_{Mg}=2\times0,1=0,2\left(mol\right)\)
\(n_{MgO}=\dfrac{2}{40}=0,05\left(mol\right)\)
Theo PT2: \(n_{HCl}=2n_{MgO}=2\times0,05=0,1\left(mol\right)\)
\(\Rightarrow\Sigma n_{HCl}=0,1+0,2=0,3\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,3}{0,3}=1\left(M\right)\)
Theo PT1,2: \(\Sigma n_{MgCl_2}=\dfrac{1}{2}\Sigma n_{HCl}=\dfrac{1}{2}\times0,3=0,15\left(mol\right)\)
\(\Rightarrow C_{M_{MgCl_2}}=\dfrac{0,15}{0,3}=0,5\left(M\right)\)
Mg + 2HCl → MgCl2 + H2 (1)
MgO + 2HCl → MgCl2 + H2O (2)
nH2=2,2422,4=0,1(mol)nH2=2,2422,4=0,1(mol)
Theo PT1: nMg=nH2=0,1(mol)nMg=nH2=0,1(mol)
⇒mMg=0,1×24=2,4(mol)⇒mMg=0,1×24=2,4(mol)
⇒mMgO=4,4−2,4=2(g)⇒mMgO=4,4−2,4=2(g)
⇒%Mg=2,44,4×100%=54,55%⇒%Mg=2,44,4×100%=54,55%
%MgO=24,4×100%=45,45%%MgO=24,4×100%=45,45%
Theo PT1: nHCl=2nMg=2×0,1=0,2(mol)nHCl=2nMg=2×0,1=0,2(mol)
nMgO=240=0,05(mol)nMgO=240=0,05(mol)
Theo PT2: nHCl=2nMgO=2×0,05=0,1(mol)nHCl=2nMgO=2×0,05=0,1(mol)
⇒ΣnHCl=0,1+0,2=0,3(mol)⇒ΣnHCl=0,1+0,2=0,3(mol)
⇒CMHCl=0,30,3=1(M)⇒CMHCl=0,30,3=1(M)
Theo PT1,2: ΣnMgCl2=12ΣnHCl=12×0,3=0,15(mol)ΣnMgCl2=12ΣnHCl=12×0,3=0,15(mol)
⇒CMMgCl2=0,150,3=0,5(M)