Cho 200 gam dd H2SO4 9,8% phản ứng với 200 gam dd KOH 5,6%. Tính C% các chất thu được sau phản ứng
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
mH2SO4=200.9,8%=19,6(g) -> nH2SO4=19,6/98=0,2(mol)
mKOH=200.5,6%=11,2(g) -> nKOH=11,2/56=0,2(mol)
PTHH: 2 KOH + H2SO4 -> K2SO4 + 2 H2O
Ta có: 0,2/2 < 0,2/1
=> H2SO4 dư, KOH hết, tính theo nKOH
=> Chất có trong dd thu được sau p.ứ: K2SO4 và H2SO4(dư)
nH2SO4(p.ứ)=nK2SO4=1/2. nKOH=1/2. 0,2=0,1(mol)
nH2SO4(dư)=0,2-0,1=0,1(mol) => mH2SO4(dư)=0,1.98=9,8(g)
mK2SO4=174. 0,1=17,4(g)
mddsau= mddH2SO4 + mddKOH= 200+200=400(g)
=>C%ddH2SO4(dư)= (9,8/400).100=2,45%
C%ddK2SO4=(17,4/400).100=4,35%
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
Ta có: \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,4\left(mol\right)\\n_{H_2}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{ddHCl}=\dfrac{0,4\cdot36,5}{14,6\%}=100\left(g\right)\\V_{H_2}=0,2\cdot22,4=4,48\left(l\right)\end{matrix}\right.\)
Bài 2:
PTHH: \(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)
Ta có: \(\left\{{}\begin{matrix}n_{KOH}=\dfrac{100\cdot11,2\%}{56}=0,2\left(mol\right)\\n_{H_2SO_4}=\dfrac{150\cdot9,8\%}{98}=0,15\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,15}{1}\) \(\Rightarrow\) H2SO4 còn dư, KOH p/ứ hết
\(\Rightarrow\left\{{}\begin{matrix}n_{K_2SO_4}=0,1\left(mol\right)\\n_{H_2SO_4\left(dư\right)}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{K_2SO_4}=0,1\cdot174=17,4\left(g\right)\\m_{H_2SO_4\left(dư\right)}=0,05\cdot98=4,9\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{ddKOH}+m_{ddH_2SO_4}=250\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{K_2SO_4}=\dfrac{17,4}{250}\cdot100\%=6,96\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{4,9}{250}\cdot100\%=1,96\%\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(m_{H_2SO_4}=\dfrac{19,6.100}{100}=19,6\left(g\right)\\ \rightarrow n_{H_2SO_4}=\dfrac{19,6}{98}=0,2\left(mol\right)\)
PTHH: 2KOH + H2SO4 ---> K2SO4 + 2H2O
0,4<-----0,2--------->0,2
\(\rightarrow m_{ddKOH}=\dfrac{0,4.56}{5,6\%}=400\left(g\right)\\ m_{dd\left(sau.pư\right)}=400+100=500\left(g\right)\\ m_{K_2SO_4}=174.0,2=34,8\left(g\right)\\ \rightarrow C\%_{K_2SO_4}=\dfrac{34,8}{500}.100\%=6,96\%\)
\(n_{H_2SO_4}=\dfrac{100.19,6\%}{98}=0,2mol\)
\(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)
0,4 0,2 0,2 ( mol )
\(m_{ddKOH}=\dfrac{0,4.56}{5,6\%}=400g\)
\(C\%_{K_2SO_4}=\dfrac{0,2.174}{100+400}.100=6,96\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,n_{HCl}=3.0,2=0,6(mol)\\ PTHH:X(OH)_n+nHCl\to XCl_n+nH_2O\\ \Rightarrow n.n_{X(OH)_n}=n_{HCl}=0,6(mol)\\ \Rightarrow M_{X(OH)_n}=\dfrac{15,6n}{0,6}=26n\\ \Rightarrow M_X+17n=26n\\ \Rightarrow M_X=9n\)
Thay \(m=3\Rightarrow M_X=27(g/mol)\)
Vậy X là nhôm (Al) và CT của bazơ là \(Al(OH)_3\)
\(b,n_{Al(OH)_3}=\dfrac{15,6}{78}=0,2(mol)\\ n_{H_2SO_4}=\dfrac{196.20\%}{100\%.98}=0,4(mol)\\ PTHH:2Al(OH)_3+3H_2SO_4\to Al_2(SO_4)_3+6H_2O\)
Vì \(\dfrac{n_{Al(OH)_3}}{2}<\dfrac{n_{H_2SO_4}}{3}\) nên \(H_2SO_4\) dư
\(\Rightarrow n_{Al_2(SO_4)_3}=\dfrac{1}{2}n_{Al(OH)_3}=0,1(mol)\\ \Rightarrow C\%_{Al_2(SO_4)_3}=\dfrac{0,1.342}{15,6+196}.100\%=16,16\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{KOH}=1.0,175=0,175\left(mol\right)\)
\(n_{H_3PO_4}=0,5.0,2=0,1\left(mol\right)\)
PTHH: KOH + H3PO4 --> KH2PO4 + H2O
______0,1<------0,1--------->0,1
KH2PO4 + KOH --> K2HPO4 + H2O
0,075<----0,075---->0,075
=> \(\left\{{}\begin{matrix}n_{KH_2PO_4}=0,025\left(mol\right)=>m_{KH_2PO_4}=0,025.136=3,4\left(g\right)\\n_{K_2HPO_4}=0,075\left(mol\right)=>m_{K_2HPO_4}=0,075.174=13,05\left(g\right)\end{matrix}\right.\)
=> mrắn = 3,4 + 13,05 = 16,45(g)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)
PT: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
a, \(n_{HCl}=2n_{CuO}=0,2\left(mol\right)\Rightarrow C\%_{HCl}=\dfrac{0,2.36,5}{200}.100\%=3,65\%\)
b, \(n_{CuCl_2}=n_{CuO}=0,1\left(mol\right)\)
\(\Rightarrow C\%_{CuCl_2}=\dfrac{0,1.135}{8+200}.100\%\approx6,49\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đáp án D
Ta có: nAxit glutamic = 0,09 mol, nHCl = 0,2 mol
⇒ ∑nCOOH + H+ = 0,09×2 + 0,2 = 0,38 mol.
+ nNaOH = 0,34 mol < ∑nCOOH + H+ = 0,38 mol ⇒ nH2O tạo thành = 0,38 mol.
Bảo toàn khối lượng ta có:
mChất rắn = 13,23 + 0,2×36,5 + 0,4×40 – 0,38×18 = 29,69 gam
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Na_2CO_3}=\dfrac{360.21,2\%}{100\%.106}=0,72(mol)\\ n_{H_2SO_4}=2,5.0,2=0,5(mol)\\ PTHH:Na_2CO_3+H_2SO_4\to Na_2SO_4+H_2O+CO_2\uparrow\\ a,\text {Vì }\dfrac{n_{Na_2CO_3}}{1}>\dfrac{n_{H_2SO_4}}{1} \text {nên }Na_2CO_3\text { dư}\\ \Rightarrow n_{CO_2}=n_{H_2SO_4}=0,5(mol)\\ \Rightarrow V_{CO_2}=0,5.22,4=11,2(l)\\\)
\(b,A:Na_2SO_4\\ n_{Na_2SO_4}=n_{H_2SO_4}=0,5(mol)\\ m_{dd_{H_2SO_4}}=200.1,1=220(g);V_{dd_{Na_2CO_3}}=\dfrac{360}{1,2}=300(ml)=0,3(l)\\ \Rightarrow C\%_{Na_2SO_4}=\dfrac{0,5.142}{360+200-0,5.44}.100\%=13,2\%\\ C_{M_{Na_2SO_4}}=\dfrac{0,5}{0,3+0,2}=1M\)
\(m_{H_2SO_4}=19,6\left(g\right)\Rightarrow n_{H_2SO_4}=0,2\left(mol\right)\)
\(m_{KOH}=11,2\left(g\right)\Rightarrow n_{KOH}=0,2\left(mol\right)\)
PTHH: \(H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O\)
\(\Rightarrow H_2SO_4\) dư.
\(\Rightarrow n_{H_2SO_4dư}=0,3\left(mol\right);n_{K_2SO_4}=0,1\left(mol\right)\)
\(\Rightarrow C\%_{H_2SO_4dư}=\dfrac{0,3.98}{400}.100\%=7,35\%\)
\(C\%_{K_2SO_4}=\dfrac{0,1.174}{400}.100\%=4,35\%\)