a) Al2O3+6HCl\(\rightarrow\) 2AlCl3 + 3H2O

MgO + 2HCl\(\rightarrow\) MgCl2 +H20

b)

\(\text{nHCl=0,5 mol}\)

a= mAl2O3+m MgO=\(\frac{\text{0,375}}{6}\).102+ \(\frac{0,125}{2}\).40=8,875g

%mAl2O3=\(\frac{\text{6,375}}{8,875}\).100%=71,83%

%mMgO=100%- 71,83%=28,17%

Mg +2HCl => MgCl2+ H2 [1]

MgO+2HCl=>MgCl2+H2O [2]

ta có nH2 =2,24:22,4=0,1 mol

theo pthh 1 ;nH2=nMg =0,1 mol

=>mMg=0,1.24 =2,4 g

=>mMgO =6-2,4=3,6g

nMgO =3,6:40=0,09 mol

%mMgO =3,6:6.100=60%

b,theo othh ;1 va 2 ta co nHCl =2nMg =0,2 mol

nHCl =2nMgO =0,18 mol

mHCl=[0,18+0,2].36,5=13,87 g

VHCl=13,87:1.1=12,6ml

1. Fe+2HCl-> FeCl2 + H2 (1)

CuO + 2HCl -> CuCl2 + H2O(2)

a.nH2=0,02(mol)=>nFe=0,02(mol)=> mFe=1,12(g)

=> nHCl(1)=0,04(mol)

nHCl dùng=\(\dfrac{15\cdot14,6}{100}=2,19\left(g\right)\)

=> nHCl=0,06(mol)

=>nHCl(2)=0,02(mol)=> nCuO=0,01(mol)

=> mCuO=0,8(g)

b. mddA=1,12+0,8+15-0,02*2=16,88(g)

mFeCl2=0,02*127=2,54(g)

mCuCl2=0,01*135=1,35(g)

C% FeCl2=\(\dfrac{2,54\cdot100}{16,88}=15,047\%\)

C% CuCl2=\(\dfrac{1,35\cdot100}{16,88}=7,9976\%\)

Đề câu 2 sao khỏi làm đi

ta có 200cm3=0,2 lítn hcl=2*0,2=0,4 molgọi số mol của caco3 là a,na2co3 là bcaco3 + 2hcl -> cacl2 + co2 + h2oa(mol)---2a(mol)--a-------a--------ana2co3 + 2hcl -> 2nacl + co2 + h2ob(mol)---2b(mol)---2b-------b------bta có100a+106b=20,62a+2b=0,4=> a=b=0,1 mol=> m caco3=10g; m na2co3=10,6 g 

PTHH: \(Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\uparrow\)  (1)

            \(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\uparrow\)  (2)

a) Ta có: \(\Sigma n_{HCl}=0,2\cdot2=0,4\left(mol\right)\)

Gọi số mol của Na₂CO₃ là \(a\) \(\Rightarrow n_{HCl\left(1\right)}=2a\left(mol\right)\)

Gọi số mol của CaCO₃ là \(b\) \(\Rightarrow n_{HCl\left(2\right)}=2b\left(mol\right)\)

Ta lập được hệ phương trình:

\(\left\{{}\begin{matrix}2a+2b=0,4\\106a+100b=20,6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{CaCO_3}=0,1\cdot100=10\left(g\right)\\m_{Na_2CO_3}=10,6\left(g\right)\end{matrix}\right.\)

b) Theo PTHH: \(\left\{{}\begin{matrix}n_{NaCl}=2n_{Na_2CO_3}=0,2mol\\n_{CaCl_2}=n_{CaCO_3}=0,1mol\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{NaCl}=0,2\cdot58,5=11,7\left(g\right)\\m_{CaCl_2}=0,1\cdot111=11,1\left(g\right)\end{matrix}\right.\)

Mặt khác: \(\left\{{}\begin{matrix}m_{CO_2}=0,2\cdot44=8,8\left(g\right)\\m_{ddHCl}=200\cdot1,2=240\left(g\right)\end{matrix}\right.\)

\(\Rightarrow m_{dd\left(saup/ư\right)}=m_{hh}+m_{ddHCl}-m_{CO_2}=251,8\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{NaCl}=\dfrac{11,7}{251,8}\cdot100\%\approx4,65\%\\C\%_{CaCl_2}=\dfrac{11,1}{251,8}\cdot100\%\approx4,41\%\end{matrix}\right.\)

MgO + 2HCl\(\rightarrow\)MgCl2 + H2O

CuO + 2HCl \(\rightarrow\)CuCl2 + H2O

Gọi số mol MgO là x, CuO là y \(\rightarrow\) 40x+80y=24 gam

Ta có: nHCl=0,3.1,6=0,48 mol =2x+2y

Giải ra nghiệm âm nhé

(Bạn xem thử lại đề)

Câu b ta có các chất là MgCl2 x mol và CuCl2 y mol

\(\rightarrow\) \(\text{V dung dịch =300ml=0,3 lít}\)

CM MgCL2=\(\frac{x}{0,3}\); CM CuCl2=\(\frac{y}{0,3}\)

Đề là a g hh nha bạn

Mg+2HCl--->MgCl2+H2

MgO+2HCl---->MgCl2+H2O

Ta có

n H2=2,24/22,4=0,1(mol)

Theo pthh1

n Mg=n H2=0,1(mol)

m Mg=0,1.24=2,4(g)

n MgCl2 ở pt1 =n H2 =0,1(mol)

m MgCl2=0,1.95=9,5(g)

m muối ở pt2 =28,5-9,5=19(g)

n MgCl2=19.95=0,2(mol)

Theo pthh2

n MgO=n MgCl2=0,2(mol0

m MgO=0,2.40=8(g)

a=m hh=8+2,4=10,4(g)

Theo pthh1

n HCl=2n H2=0,2(mol)

Theo pthh2

n HCl=2n MgCl2=0,4(mol)

\(\sum nHCl=0,2+0,4=0,6\left(mol\right)\)

x=0,6/0,4=1,5(M)

Ngắn gọn hơn bài Dương nhé:

Mg+2HCl\(\rightarrow\)MgCl2+H2 (1)

MgO+2HCl\(\rightarrow\)MgCl2+H2O (2)

nH2=\(\frac{2,24}{22,4}\)=0,1 mol

\(\rightarrow\) nMg=0,1 mol, nMgCl2 (1)=0,1 mol

\(\rightarrow\)mMgCl2(1)=0,1.95=9,5g

\(\rightarrow\) mMgCl2(2)=28,5-9,5=19g

\(\rightarrow\) nMgCl2(2)=nMgO=19:95=0,2 mol

Ta có:

x gam hh tpu= mMg+mMgO=0,1.24+0,2.95=21,4g

CM HCl=(0,1.2+0,2.2)/0,4=1,5M

a(g) là klg hh

PTHH1:Mg+2HCl->MgCl2+H2

mol:.....0,1........0,2.......0,1.....0,1

PTHH2:MgO+2HCl->MgCl2+H2O

mol:0,2...........0,4.............0,2.....0,2

mH2=2,24:22,4=0,1(mol)

mMgCl2(pt1)=0,1.95=9,5(g)

mMgCl2(pt2)=28,5-9,5=19(g)

nMgCl(pt2)=19:95=0,2(mol)

a=0,1.24+0,2.40=10,4(g)

CM\(_{HCl}\)=\(\frac{0,1.2+0,2.2}{0,4}\)=1,5M

Câu 1:

Gọi : nMg=a(mol); nMgO=b(mol) (a,b>0)

a) PTHH: Mg + 2 HCl -> MgCl2 + H2

a________2a_______a______a(mol)

MgO +2 HCl -> MgCl2 + H2O

b_____2b_______b___b(mol)

Ta có hpt:

\(\left\{{}\begin{matrix}24a+40b=8,8\\22,4a=4,48\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\)

=> mMg=0,2.24=4,8(g)

=>%mMg= (4,8/8,8).100=54,545%

=> %mMgO= 45,455%

b) m(muối)=mMg2+ + mCl- = 0,3. 24 + 0,6.35,5=28,5(g)

c) V=VddHCl=(2a+2b)/2=0,3(l)=300(ml)

Câu 2:

Ta có: \(\left\{{}\begin{matrix}n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\n_{HCl}=0,4\cdot2=0,8\left(mol\right)\end{matrix}\right.\)

PTHH: \(Ca+2HCl\rightarrow CaCl_2+H_2\uparrow\)

               0,2____0,4_____0,2____0,2   (mol)

           \(CaO+2HCl\rightarrow CaCl_2+H_2O\)

                0,2____0,4______0,2____0,2  (mol)

Ta có: \(\left\{{}\begin{matrix}\%m_{Ca}=\dfrac{0,2\cdot40}{0,2\cdot40+0,2\cdot56}\cdot100\%\approx41,67\%\\\%m_{CaO}=58,33\%\\m_{CaCl_2}=\left(0,2+0,2\right)\cdot111=44,4\left(g\right)\end{matrix}\right.\)

nH2SO4=\(\frac{\text{200.9,8%}}{98}\)=0,2

nH2=0,15

a. Mg+H2SO4\(\rightarrow\)MgSO4+H2 (1)

CaO+H2SO4\(\rightarrow\)CaSO4+H2O (2)

\(\text{nMg=nH2=0,15=nH2SO4(1)=nMgSO4}\)

\(\rightarrow\)nH2SO4(2)=0,05

\(\rightarrow\)nCaO=0,05=nCaSO4

\(\rightarrow\)m hỗn hợp muối sau phản ứng=\(\text{0,15.120+0,05.136=24,8}\)

b. mdd sau phản ứng=\(\text{0,15.24+0,05.56+200-0,15.2=206,1}\)

c%MgSO4=0,15.120/206,1=8,73%