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\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\
pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,1 0,2 0,1 0,1
\(m_{HCl}=0,2.36,5=7,3\left(g\right)\\
V_{H_2}=0,1.22,4=2,24l\\
m_{\text{dd}}=6,5+200-\left(0,1.2\right)=206,3g\)
bài 2 :
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\
pthh:Mg+2HCl\rightarrow MgCl_2+H_2\)
0,2 0,4 0,2 0,2
\(m_{HCl}=0,4.36,5=14,6g\\
V_{H_2}=0,2.22,4=4,48l\\
m\text{dd}=4,8+200-0,4=204,4g\\
C\%=\dfrac{0,2.136}{204,4}.100\%=13,3\%\)
Bài 1: nZn=m/M=19,5/65=0,3(mol)
PT:
Zn + 2HCl -> ZnCl2 +H2\(\uparrow\)
1........2..............1..........1. (mol)
0,3->0,6 -> 0,3 -> 0,3 (mol)
VH2=n.22,4=0,3.22,4=6,72(lít)
b)PT:
3H2 + Fe2O3 \(\underrightarrow{t^0}\) 2Fe + 3H2O
3............1...........2...............3 (mol)
0,3 -> 0,1 -> 0,2 -> 0,3 (mol)
=> mFe=n.M=0,2.56=11,2(g)
nMg=m/M=2,4/24=0,1(mol)
PT:
Mg + 2HCl -> MgCl2 + H2\(\uparrow\)
1.............2............1.............1 (mol)
0,1-> 0,2 -> 0,1 -> 0,1 (mol)
VH2=n.22,4=0,1.22,4=2,24(lít
b) mHCl=n.M=0,2.36,5=7,3(g)
=> md d HCl=\(\dfrac{m_{HCl}.100\%}{C\%}=\dfrac{7,3.100}{20}=36,5\left(g\right)\)
c) mMgCl2=n.M=0,1.95=9,5(g)
=> md d sau phản ứng= mMg + mHCl-mH2=2,4 + 36,5-(0,1.2)=38,7(g)
=> C% d d sau khi trộn=\(\dfrac{m_{MgCl_2}.100\%}{m_{ddsauphanung}}=\dfrac{9,5.100}{38,7}\approx24,55\left(\%\right)\)
VHCl=D.m=1,1.36,5=40,15(ml)=0,04015(lít)
=> CM HCl=\(\dfrac{n}{V}=\dfrac{0,2}{0,04015}\approx5\left(M\right)\)
nMg = \(\dfrac{2,4}{24}=0,1\) mol
Pt: Mg + .....2HCl --> MgCl2 + H2
0,1 mol-> 0,2 mol-> 0,1 mol-> 0,1 mol
VH2 = 0,1 . 22,4 = 2,24 (lít)
mdd HCl pứ = \(\dfrac{0,2\times36,5\times100}{20}=36,5\left(g\right)\)
mdd sau pứ = mMg + mdd HCl pứ - mH2 = 2,4 + 36,5 - 0,1 . 2 = 38,7 (g)
C% dd MgCl2 = \(\dfrac{0,1\times95}{38,7}.100\%=24,55\%\)
nMg=2,4/24=0,1(mol)
Mg+2HCl--->MgCl2+H2
0,1____0,2_____0,1__0,1
VH2=0,1.22,4=2,24(l)
b)mHCl=0,2.36,5=7,3(g)
=>mddHCl=7,3.100/20=36,5(g)
c)mdd=2,4+36,5-0,1.2=38,7(g)
mMgCl2=0,1.95=9,5(g)
=>C%MgCl2=9,5/38,7.100%~24,55%
Fe + 2HCl -> FeCl2 + H2
nFe = 5,6/56 = 0,1 mol
=>nH2 = 0,1 mol
=> VH2= 0,1*22,4= 2,24 lít
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
PTHH: Fe + 2HCl ---> FeCl2 + H2
0,1-->0,2------------------>0,1
=> \(\left\{{}\begin{matrix}m_{ddHCl}=\dfrac{0,2.36,5}{15\%}=\dfrac{146}{3}\left(g\right)\\V_{H_2}=0,1.22,4=4,48\left(l\right)\end{matrix}\right.\)
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: \(m_{dd}=m_{ddKOH}+m_{ddH_2SO_4}=250\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{K_2SO_4}=\dfrac{17,4}{250}\cdot100\%=6,96\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{4,9}{250}\cdot100\%=1,96\%\end{matrix}\right.\)
Đầu tiên bạn tính n H2 = cách bảo toàn e =» n hcl pư =» m dd hcl pư
Bạn bảo toàn ntố Fe để tím n FeCl2 =» m FeCl2 (dd B)
C% dd B = m FeCl 2 / (m Fe + m dd HCl)
a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(n_{HCl}=0,15.4=0,6\left(mol\right)\)
Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{HCl}=0,3\left(mol\right)\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)
b, \(n_{Zn}=\dfrac{1}{2}n_{HCl}=0,3\left(mol\right)\Rightarrow m_{Zn}=0,3.65=19,5\left(g\right)\)
c, \(n_{ZnCl_2}=\dfrac{1}{2}n_{HCl}=0,3\left(mol\right)\Rightarrow C_{M_{ZnCl_2}}=\dfrac{0,3}{0,15}=2\left(M\right)\)
\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
a.
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(V_{H_2}=24,79.0,2=4,958\left(l\right)\)
b.
\(n_{HCl}=2.n_{Fe}=0,4\left(mol\right)\\ CM_{HCl}=\dfrac{0,4}{0,2}=2M\)
\(PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\)
Ta có:
\(n_{Mg}=\frac{2,4}{24}=0,1\left(mol\right)\)
\(\Rightarrow n_{H2}=n_{Mg}=0,1\left(mol\right)\Rightarrow V_{H2}=0,1.22,4=2,24\left(l\right)\)
\(\Rightarrow n_{HCl}=2n_{H2}=0,2\left(mol\right)\Rightarrow m_{HCl}=0,2.36,5=7,3\left(g\right)\)
Do HCl chiếm 20% khối lượng dung dịch
\(\Rightarrow m_{HCl}=\frac{7,3.100}{20}=36,5\left(g\right)\)
\(\Rightarrow n_{MgCl2}=n_{Mg}=0,1\left(mol\right)\Rightarrow m_{MgCl2}=0,1.\left(24+35,5.2\right)=9,5\left(g\right)\)
\(m_{dd\left(spu\right)}=38,7\left(g\right)\Rightarrow C\%=\frac{9,5}{38,7}.100\%=24,55\%\)
Ta co pthh
Mg + 2HCl \(\rightarrow\) MgCl2 + H2
Theo de bai ta co
nMg=\(\dfrac{2,4}{24}=0,1\left(mol\right)\)
a,Theo pthh
nH2=nMg=0,1 mol
\(\Rightarrow\) VH2\(_{\left(dktc\right)}=0,1.22,4=2,24\left(l\right)\)
b, Theo pthh
nHCl=2nMg=2.0,1=0,2 mol
\(\Rightarrow mct=mHCl=0,2.36,5=7,3\left(g\right)\)
\(\Rightarrow mdd_{HCl}=\dfrac{mct.100\%}{C\%}=\dfrac{7,3.100\%}{20\%}=36,5\left(g\right)\)
c, Theo Pthh ta co
nMgCl2=nMg=0,1 mol
Theo de bai ta co
V\(dd_{HCl}=\dfrac{mdd}{D_{dd}}=\dfrac{36,5}{1,1}\approx33,18ml=0,3318l\)
\(\Rightarrow CM=\dfrac{n}{V}=\dfrac{0,1}{0,3318}\approx0,30M\)
nMg=m/M=2,4/24=0,1(mol)
PT:
Mg + 2HCl -> MgCl2 +H2
1..........2................1.........1 (mol)
0,1 -> 0,2 -> 0,1 -> 0,1 (mol)
Khí thu được là H2
VH2=n.22,4=0,1.22,4=2,24(lít)
b) mHCl=n.M=0,2.36,5=7,3(gam)
=> \(m_{ddHCl}=\dfrac{m_{HCl}.100\%}{C\%}=\dfrac{7,3.100}{20}=36,5\left(g\right)\)
c) md d sau phản ứng=mMg + mHCl-mH2=2,4+36,5-(0,1.2)=38,7(g)
mMgCl2=n.M=0,1.95=9,5(g)
\(\Rightarrow C\%_{ddsauphanung}=\dfrac{m_{MgCl_2}.100\%}{m_{ddsauphanung}}=\dfrac{9,5.100}{38,7}\approx24,55\left(\%\right)\)
Ta có: \(V_{ddHCl}=\dfrac{m_{ddHCl}}{D}=\dfrac{36,5}{1,1}\approx33,18\left(ml\right)\)= 0,3318(lít)
=> CM=\(\dfrac{n}{V}=\dfrac{0,1}{0,3318}\approx0,3\left(M\right)\)