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Câu 2:
\(n_{MgBr_2}=\dfrac{14,72}{184}=0,08\left(mol\right)\\ Mg+Br_2\rightarrow MgBr_2\\ n_{Mg}=n_{Br_2}=n_{MgBr_2}=0,08\left(mol\right)\\ a=m_{Mg}=24.0,08=1,92\left(g\right)\\ m_{Br_2}=160.0,08=12,8\left(g\right)\)
Câu 1:
\(n_{AlBr_3}=\dfrac{106,8}{267}=0,4\left(mol\right)\\ 2Al+3Br_2\rightarrow2AlBr_3\\ n_{Al}=n_{AlBr_3}=0,4\left(mol\right)\\ n_{Br_2}=\dfrac{3}{2}.0,4=0,6\left(mol\right)\\ a=m_{Al}=0,4.27=10,8\left(g\right)\\ m_{Br_2}=160.0,6=96\left(g\right)\)
Câu 2 :
\(n_{Cu}=\dfrac{22,4}{64}=0,35\left(mol\right)\)
Pt : \(Cu+Cl_2\underrightarrow{t^o}CuCl_2|\)
1 1 1
0,35 0,35 0,35
\(n_{CuCl2}=\dfrac{0,35.1}{1}=0,35\left(mol\right)\)
⇒ \(m_{CuCl2}=0,35.135=47,25\left(g\right)\)
\(n_{Cl2}=\dfrac{0,35.1}{1}=0,35\left(mol\right)\)
\(V_{Cl2\left(dtkc\right)}=0,35.22,4=7,84\left(l\right)\)
Chúc bạn học tốt
\(Câu4\\ n_{Cl_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ 2Al+3Cl_2\rightarrow\left(t^o\right)2AlCl_3\\ n_{Al}=n_{AlCl_3}=\dfrac{2}{3}.0,15=0,1\left(mol\right)\\ \Rightarrow m=m_{Al}=0,1.27=2,7\left(g\right)\\ m_{AlCl_3}=133,5.0,1=13,35\left(g\right)\)
Câu 1:
\(2Na+Br_2\rightarrow2NaBr\\ n_{NaBr}=\dfrac{61,8}{103}=0,6\left(mol\right)\\ n_{Na}=n_{NaBr}=0,6\left(mol\right)\\ n_{Br_2}=\dfrac{0,6}{2}=0,3\left(mol\right)\\ \Rightarrow a=m_{Na}=0,6.23=13,8\left(g\right)\\ m_{Br_2}=0,3.160=48\left(g\right)\\ m_{ddBr_2}=\dfrac{48}{5\%}=960\left(g\right)\)
Câu 2:
\(2Fe+3Cl_2\rightarrow\left(t^o\right)2FeCl_3\\ n_{FeCl_3}=\dfrac{40,625}{162,5}=0,25\left(mol\right)\\ n_{Fe}=n_{FeCl_3}=0,25\left(mol\right)\\ \Rightarrow m=m_{Fe}=0,25.56=14\left(g\right)\\ n_{Cl_2}=\dfrac{3}{2}.0,25=0,375\left(mol\right)\\ V_{Cl_2\left(đktc\right)}=0,375.22,4=8,4\left(l\right)\)
Câu 1:
\(Mg+Br_2\rightarrow MgBr_2\\ n_{Br_2}=\dfrac{11,2}{160}=0,07\left(mol\right)=n_{Mg}=n_{MgBr_2}\\ a=m_{Mg}=0,07.24=1,68\left(g\right)\\ m_{MgBr_2}=184.0,07=12,88\left(g\right)\)
TH1:
\(2R+2nH_2SO_4\rightarrow R_2\left(SO_4\right)_n+nSO_2+2nH_2O\)
\(\frac{n_R}{n_{H2SO4}}=\frac{1}{1,25}\Rightarrow\frac{1}{n}=\frac{1}{1,25}\)
\(\Rightarrow n=1,25\)
\(\Rightarrow2R+2,5H_2SO_4\rightarrow R_2\left(SO_4\right)_{1,25}+1,25SO_2+2,5H_2O\)
\(n_{Br2}=0,1\left(mol\right)\)
\(SO_2+Br_2+2H_2O\rightarrow2HBr+H_2SO_4\)
0,1____0,1______________________________
\(n_{R2\left(SO4\right)1,5}=0,08\left(mol\right)\)
\(\Rightarrow M_{R2\left(SO4\right)1,5}=\frac{12}{0,08}=150=2R+1,5.96\)
\(\Rightarrow R=3\) (loại)
TH2 : \(8R+5nH_2SO_4\rightarrow4R_2\left(SO_4\right)_n+nH_2S+4nH_2O\)
\(\frac{n_2}{n_{H2SO4}}=\frac{1}{1,25}=\frac{4}{5}\Rightarrow\frac{8}{5n}=\frac{4}{5}\)
\(\Rightarrow n=2\)
\(4R+5H_2SO_4\rightarrow4RSO_4+H_2S+4H_2O\)
\(\Rightarrow n_{Br2}=0,1\left(mol\right)\)
\(H_2S+4Br_2+4H_2O\rightarrow8HBr+H_2SO_4\)
0,025___0,1___________________________
\(\Rightarrow n_{RSO4}=0,1\left(mol\right)\)
\(\Rightarrow M_{RSO4}=\frac{12}{0,1}=120=R+96\)
\(\Rightarrow R=24\left(Mg\right)\)
Vậy kim loại R là Magie (Mg)
\(a,PTHH:R+2AgNO_3\to R(NO_3)_2+2Ag\\ \Rightarrow n_{R}=n_{R(NO_3)_2}\\ \Rightarrow \dfrac{2,8}{M_R}=\dfrac{9}{M_R+124}\\ \Rightarrow M_R=56(g/mol)\)
Vậy R là sắt (Fe)
\(b,n_{R}=\dfrac{2,8}{56}=0,05(mol)\\ \Rightarrow n_{AgNO_3}=0,1(mol)\\ \Rightarrow m_{dd_{AgNO_3}}=\dfrac{0,1.170}{5\%}=340(g)\\ c,n_{Fe(NO_3)_2}=n_{Fe}=0,05(mol);n_{Ag}=0,1(mol)\\ \Rightarrow C\%_{Fe(NO_3)_2}=\dfrac{0,05.180}{2,8+340-0,1.108}.100\%=2,71\%\)
\(Mg+CuSO_4--->MgSO_4+Cu\left(1\right)\)
0,1_______0,1__________0,1_________0,1
\(Zn+CuSO_4--->ZnSO_4+Cu\left(2\right)\)
0,15______0,15__________0,15_____0,15
Do sau pứ thu đc \(d^2B\) gồm 2 muối và 19,25 gam hỗn hợp kim loại
nên Mg pứ hết , Zn pứ 1 phần ,CuSO4 pứ hết
\(n_{CuSO_4}=0,2.1,25=0,25\left(g\right)\)
a) Đặt a ,b, c lần lượt là số mol của Mg , Zn pứ , Zn dư
Theo đề ra ta có :
\(24a+65b+65c=15,4\left(I\right)\)
\(a+b=0,25\left(II\right)\)
\(64.\left(a+b\right)+65c=19,25\left(III\right)\)
=> a=0,1
b=0,15
c=0,05
=> \(\%m_{Mg}=\frac{24.0,1}{15,4}.100=15,58\%\)
=> \(\%m_{Zn}=100-15,58=84,42\%\)
b) \(m_{d^2CuSO_4}=0,25.160=40\left(g\right)\)
=> \(m_{d^2sau}=40-0,6=39,4\left(g\right)\)
=> \(C\%_{MgSO_4}=\frac{0,1.120}{39,4}.100=30,45\%\)
=>\(C\%_{ZnSO_4}=\frac{0,15.161}{39,4}.100=61,3\%\)
c) \(2NaOH+MgSO_4-->Na_2SO_4+Mg\left(OH\right)_2\downarrow\left(3\right)\)
0,2_________0,1
\(2NaOH+ZnSO_4-->Na_2SO_4+Zn\left(OH\right)_2\downarrow\left(4\right)\)
0,3_________0,15_____________________0,15
\(2NaOH+Zn\left(OH\right)_2-->Na_2ZnO_2+2H_2O\left(5\right)\)
0,3_______0,15
Khi cho B pứ vừa đủ với d2 NaOH để thu được ↓ lớn nhất ⇔ \(Zn\left(OH\right)_2\) không tan
=> \(n_{NaOH}=0,3+0,2=0,5\left(mol\right)\)
=> V=?
d) Khi cho B pứ vừa đủ với d2 NaOH để thu được ↓ nhỏ nhất ⇔ \(Zn\left(OH\right)_2\) tan hết
=> \(n_{NaOH}=0,2+0,3+0,2=0,7\left(mol\right)\)
=> V=?
(Đề hình như thiếu \(C_M\) )
Br2 + 2NaI \(\rightarrow\) 2NaBr + I2
0,5x........x.............x
m NaI = 150x (g)
m NaBr = 103x (g)
m NaI - m NaBr2 = 2,82 (g)
\(\Rightarrow\) 150x - 103x = 2,82
\(\Rightarrow\) 47x = 2,82
\(\Rightarrow\) x = 0,06
n Br2 = 0,5 . 0,06 = 0,03 (mol)
m Br2 = 0,03 . 160 = 4,8 (g)
Mg + 2HCl \(\rightarrow\)MgCl2 + H2
nH2=\(\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Theo pTHH ta có:
nMg=nH2=0,2(mol)
mMg=24.0,2=4,8(g)
mCu=11,3-4,8=6,5(g)
b;
Theo pTHH ta có:
2nMg=nHCl=0,4(mol)
VHCl=\(\dfrac{0,4}{0,5}=0,8\left(lít\right)\)
2Al+3Br2->2AlBr3
0,3---0,45----0,3 mol
n Al=\(\dfrac{8,1}{27}\)=0,3 mol
=>mBr2=0,45.160=72g
=>m AlBr3=0,3.267=80,1g
\(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\\ 2Al+3Br_2\rightarrow2AlBr_3\\ n_{AlBr_3}=n_{Al}=0,3\left(mol\right)\\ n_{Br_2}=\dfrac{3}{2}.0,3=0,45\left(mol\right)\\ m_{AlBr_3}=267.0,3=80,1\left(g\right)\\ m_{Br_2}=0,45.160=72\left(g\right)\)