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Cho kim loại Magie tác dụng vừa đủ với 200 gam dung dịch axit axetic 15%.
a. Tính khối lượng Magie phản ứng ?
b. Tính nồng độ phần trăm dung dịch muối thu được sau phản ứng ?
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mCH3COOH= 200.15%= 30(g) => nCH3COOH= 30/60=0,5(mol)
a) Mg + 2 CH3COOH -> (CH3COO)2Mg + H2
0,25___0,5_______0,25_____________0,25(mol)
mMg= 0,25.24= 6(g)
b) m(CH3COO)2Mg=142.0,25=35,5(g)
mdd(CH3COO)2Mg= 6+200-0,25.2=205,5(g)
=> \(C\%dd\left(CH3COO\right)2Mg=\frac{35,5}{205,5}.100\approx17,275\%\)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
a) Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)
1 2 1 1
0,2 0,4 0,2 0,2
b) \(n_{HCl}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)
⇒ \(m_{HCl}=0,4.36,5=14,6\left(g\right)\)
\(C_{ddHCl}=\dfrac{14,6.100}{100}=14,6\)0/0
c) \(n_{ZnCl2}=\dfrac{0,4.1}{2}=0,2\left(mol\right)\)
⇒ \(m_{ZnCl2}=0,2.136=27,2\left(g\right)\)
\(m_{ddspu}=13+100-\left(0,2.2\right)=112,6\left(g\right)\)
\(C_{ZnCl2}=\dfrac{27,2.100}{112,6}=24,16\)0/0
Chúc bạn học tốt
\(n_{Zn}=\dfrac{13}{65}=0.2\left(mol\right)\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(0.2........0.4..........0.2.......0.2\)
\(m_{HCl}=0.4\cdot36.5=14.6\left(g\right)\)
\(C\%_{HCl}=\dfrac{14.6}{100}\cdot100\%=14.6\%\)
\(m_{ZnCl_2}=0.2\cdot136=27.2\left(g\right)\)
\(m_{\text{dung dịch sau phản ứng}}=13+100-0.2\cdot2=112.6\left(g\right)\)
\(C\%_{ZnCl_2}=\dfrac{27.2}{112.6}\cdot100\%=24.1\%\)
a) Na2CO3 + 2CH3COOH --> 2CH3COONa + CO2 + H2O
b) \(n_{CH_3COOH}=\dfrac{25.6\%}{60}=0,025\left(mol\right)\)
PTHH: Na2CO3 + 2CH3COOH --> 2CH3COONa + CO2 + H2O
0,0125<-----0,025------------>0,025------>0,0125
=> \(m_{Na_2CO_3}=0,0125.106=1,325\left(g\right)\)
c) \(m_{dd.sau.pư}=1,325+25-0,0125.44=25,775\left(g\right)\)
\(C\%_{dd.CH_3COONa}=\dfrac{0,025.82}{25,775}.100\%=7,95\%\)
\(n_{Zn}=\dfrac{13}{65}=0,2(mol)\\ a,PTHH:Zn+2HCl\to ZnCl_2+H_2\\ b,n_{HCl}=0,4(mol)\\ \Rightarrow C\%_{HCl}=\dfrac{0,4.36,5}{100}.100\%=14,6\%\\ c,n_{ZnCl_2}=n_{H_2}=0,2(mol)\\ \Rightarrow m_{ZnCl_2}=0,2.136=27,2(g)\\ \Rightarrow C\%_{ZnCl_2}=\dfrac{27,2}{13+100-0,2.2}.100\%\approx 24,16\%\)
a) MgO + 2HCl → MgCl2 + H2O (1)
b) \(n_{MgO}=\dfrac{6}{40}=0,15\left(mol\right)\)
Theo PT1: \(n_{HCl}=2n_{MgO}=2\times0,15=0,3\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,3\times36,5=10,95\left(g\right)\)
\(\Rightarrow C\%_{ddHCl}=\dfrac{10,95}{200}\times100\%=5,475\%\)
c) Theo PT1: \(n_{MgCl_2}=n_{MgO}=0,15\left(mol\right)\)
\(\Rightarrow m_{MgCl_2}=0,15\times95=14,25\left(g\right)\)
\(\Sigma m_{dd}=6+200=206\left(g\right)\)
\(\Rightarrow C\%_{ddMgCl_2}=\dfrac{14,25}{206}\times100\%=6,92\%\)
d) HCl + KOH → KCl + H2O (2)
Theo PT2: \(n_{KOH}=n_{HCl}=0,3\left(mol\right)\)
\(\Rightarrow V_{ddKOH}=\dfrac{0,3}{0,5}=0,6\left(l\right)=600\left(ml\right)\)
a, \(Zn+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2\)
b, Phần này đề hỏi tính khối lượng gì bạn nhỉ?
c, \(n_{Zn}=\dfrac{32,5}{65}=0,5\left(mol\right)\)
Theo PT: \(n_{CH_3COOH}=2n_{Zn}=1\left(mol\right)\)
\(\Rightarrow m_{ddCH_3COOH}=\dfrac{1.60}{36\%}=\dfrac{500}{3}\left(g\right)\)
Ta có:
\(n_{Zn}=\frac{13}{65}=0,2\left(mol\right)\)
\(2CH_3COOH+Zn\rightarrow\left(CH_3COO\right)_2Zn+H_2\)
\(\Rightarrow n_{CH3COOH}=2n_{Zn}=0,2.2=0,4\left(mol\right)\)
\(m_{dd\left(CH3COOH\right)}=\frac{0,4.60}{12\%}=200\left(g\right)\)
\(n_{H2}=n_{Zn}=0,2\left(mol\right)\Rightarrow m_{dd\left(spu\right)}=212,6\left(g\right)\)
\(n_{\left(CH3COO\right)2Zn}=n_{Zn}=0,2\left(mol\right)\)
\(\Rightarrow C\%_{\left(CH3COO\right)2Zn}=\frac{0,2.183}{212,6}.100\%=17,22\%\)