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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:
\(Mg+Br_2\rightarrow MgBr_2\\ n_{Br_2}=\dfrac{11,2}{160}=0,07\left(mol\right)=n_{Mg}=n_{MgBr_2}\\ a=m_{Mg}=0,07.24=1,68\left(g\right)\\ m_{MgBr_2}=184.0,07=12,88\left(g\right)\)
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:
\(n_{AlBr_3}=\dfrac{106,8}{267}=0,4\left(mol\right)\\ 2Al+3Br_2\rightarrow2AlBr_3\\ n_{Al}=n_{AlBr_3}=0,4\left(mol\right)\\ n_{Br_2}=\dfrac{3}{2}.0,4=0,6\left(mol\right)\\ a=m_{Al}=0,4.27=10,8\left(g\right)\\ m_{Br_2}=160.0,6=96\left(g\right)\)
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
\(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\\ 2Al+3Br_2\rightarrow2AlBr_3\\ n_{AlBr_3}=n_{Al}=0,3\left(mol\right)\\ n_{Br_2}=\dfrac{3}{2}.0,3=0,45\left(mol\right)\\ m_{AlBr_3}=267.0,3=80,1\left(g\right)\\ m_{Br_2}=0,45.160=72\left(g\right)\)
\(n_{Fe_2O_3}=\dfrac{8}{160}=0,05\left(mol\right)\)
PTHH: Fe2O3 + 3H2SO4 --> Fe2(SO4)3 + 3H2O
______0,05------>0,15--------->0,05
=> mH2SO4 = 0,15.98 = 14,7(g)
=> \(C\%\left(H_2SO_4\right)=\dfrac{14,7}{100}.100\%=14,7\%\)
\(C\%\left(Fe_2\left(SO_4\right)_3\right)=\dfrac{0,05.400}{8+100}.100\%=18,52\%\)
PTHH: Fe2(SO4)3 + 6NaOH --> 2Fe(OH)3\(\downarrow\) + 3Na2SO4
________0,05----------------------->0,1
=> mFe(OH)3 = 0,1.107=10,7(g)
Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
____0,1______0,2_____0,1____0,1 (mol)
a, \(C_{M_{HCl}}=\dfrac{0,2}{0,15}=\dfrac{4}{3}\left(M\right)\)
\(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
b, \(FeCl_2+2NaOH\rightarrow2NaCl+Fe\left(OH\right)_2\)
Theo PT: \(n_{NaOH}=2n_{FeCl_2}=0,2\left(mol\right)\)
\(\Rightarrow V_{NaOH}=\dfrac{0,2}{2}=0,1\left(l\right)\)
Phương trình hoá học : 2Fe + 3 Cl 2 → t ° FeCl 3
Theo định luật bảo toàn khối lượng :
m Fe + m Cl 2 = m FeCl 3
m Cl 2 = m FeCl 3 - m Fe = 16,25 - 5,6 = 10,65g
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:
\(2Fe+3Cl_2\rightarrow\left(t^o\right)2FeCl_3\\ n_{FeCl_3}=\dfrac{40,625}{162,5}=0,25\left(mol\right)\\ n_{Fe}=n_{FeCl_3}=0,25\left(mol\right)\\ \Rightarrow m=m_{Fe}=0,25.56=14\left(g\right)\\ n_{Cl_2}=\dfrac{3}{2}.0,25=0,375\left(mol\right)\\ V_{Cl_2\left(đktc\right)}=0,375.22,4=8,4\left(l\right)\)