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\(n_{KMnO_4}=\dfrac{39,5}{158}=0,25\left(mol\right)\)
\(2KMnO_4--t^0->K_2MnO_4+MnO_2+O_2\)
0,25........................................................................0,125(mol)
\(m_{90\%O}=\dfrac{0,125.32.90\%}{100\%}=3,6\left(g\right)\)
\(n_{O_2}=\dfrac{3,6}{32}=0,1125\left(mol\right)\)
\(2xR+yO_2\rightarrow2R_xO_y\)
\(\dfrac{0,225x}{y}\) ......0,1125 .......0,225
\(M_R=\dfrac{5,4}{\dfrac{0,225x}{y}}=\dfrac{24y}{x}\left(1\right)\)
Theo định luật bảo toàn khối lượng ta có
\(m_{R_xO_y}=0,225\left(xM_R+16y\right)=5,4+3,6\)
\(\Leftrightarrow\dfrac{24y}{x}.0,225x+3,6y=9\)
\(\Rightarrow y=1\)
\(\Rightarrow x.M_R=24\left(\dfrac{g}{mol}\right)\)
| x | 1 | 2 | 3 |
| MR | 24 | 12 | 8 |
| chọn | loại | chọn |
vậy R: Magie
\(n_R=\dfrac{5,4}{M_R}\)
\(n_{KMnO_4}=0,25\left(mol\right)\)
\(2KMnO_4-t^0->MnO_2+K_2MnO_4+O_2\)
\(0,25mol..................................0,125mol\)
Mà khi đốt cháy hoàn toàn 5,4 g kim loại R chỉ cần dùng một lượng 90% lượng oxi sinh ra nên : \(n_{O_2}=90\%.0,125=0,1125\left(mol\right)\)
\(4R+nO_2->2R_2O_n\)
\(\dfrac{0,45}{n}......0,1125\)
\(\dfrac{5,4}{M_R}=\dfrac{0,45}{n}\)
\(1\le n\le3\)
\(n=1=>M_R=12\left(loại\right)\)
\(n=2=M_R=24\left(Mg\right)\)
\(n=3=>M_R=36\left(loại\right)\)
Vậy R là Mg .
\(n_{KMnO_4}=\dfrac{63,2}{158}=0,4\left(mol\right)\\
pthh:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
0,4 0,2
=> \(V_{O_2\left(lt\right)}=0,2.22,4=4,48\left(l\right)\\
V_{O_2\left(tt\right)}=\dfrac{90.4,48}{100}=4,032\left(l\right)\)
a)\(n_{KMnO_4}\)=39,5:158=0,25(mol)
Ta có PTHH:
2\(KMnO_4\)\(\underrightarrow{to}\)\(K_2MnO_4+MnO_2+O_2\)(1)
......0,25..............................................0,125(mol)
Theo PTHH:\(n_{O_2\left(1\right)}\)=0,125(mol)
=>\(n_{O_2\left(cần\right)}\)=90%.0,125=0,1125(mol)
=>\(m_{O_2\left(cần\right)}\)=0,1125.32=3,6(g)
Gọi n là hóa trị của R
4R+n\(O_2\)\(\underrightarrow{to}\)2\(R_2O_n\)
4R...32n..................(g)
5,4....3,6..................(g)
Theo PTHH:3,6.4R=5,4.32n=>R=12n
Vì n là hóa trị của R nên n\(\in\){1;2;3;\(\dfrac83\)}
Biện luận:
| n | 1 | 2 | 3 | 8/3 |
| R | 12 | 24 | 36 | 32 |
=>n=2;R=24(Mg) là phù hợp
Vậy R là Mg
b)\(n_{Mg}\)=5,4:24=0,225(mol)
Ta có PTHH:
Mg+2HCl->Mg\(Cl_2\)+\(H_2\)
0,225..0,45........................(mol)
Theo PTHH:\(m_{HCl}\)=0,45.36,5=16,425(g)
mà \(C_{\%ddHCl}\)=14,6%
=>\(m_{dd\left(gt\right)}\)=16,425:14,6%=112,5(g)
mà dd lấy dư 20% nên:
=>\(m_{dd\left(cần\right)}\)=112,5+20%.112,5=135(g)
\(a) 4P+ 5O_2 \xrightarrow{t^o} 2P_2O_5\\ b) n_{O_2} = \dfrac{1,12}{22,4} = 0,05(mol)\\ n_{P_2O_5} = \dfrac{2}{5}n_{O_2} = 0,02(mol)\\ m_{P_2O_5} = 0,02.142 = 2,84(gam) c) 2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2\\ n_{KClO_3} = \dfrac{2}{3}n_{O_2} = \dfrac{0,1}{3}(mol)\\ m_{KClO_3} = \dfrac{0,1}{3}122,5 = 4,083(gam)\)
\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\a, PTHH:4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\\ b,n_{O_2}=\dfrac{3}{4}.n_{Al}=\dfrac{3.0,2}{4}=0,15\left(mol\right)\\ \Rightarrow V_{O_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\\ c,2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\\ n_{KMnO_4}=2.n_{O_2}=2.0,15=0,3\left(mol\right)\\ \Rightarrow m_{KMnO_4}=158.0,3=47,4\left(g\right)\)
áp dụng ĐLBTKL:
mR + mO2 = mR2O3
=> mO2=20,4-10,8=9,6(g)
=> nO2=9,6/32=0,3(mol)
4R + 3O2 ---to---> 2R2O3
0,4........0,3
MR=10,8/0,4=27(g)
=> R là nhôm ......Al
\(2Cu+O_2\rightarrow\left(t^o\right)2CuO\\ n_{CuO}=n_{Cu}=\dfrac{6,4}{64}=0,1\left(mol\right);n_{O_2}=\dfrac{0,1}{2}=0,05\left(mol\right)\\ a,V_{O_2\left(đktc\right)}=0,05.22,4=1,12\left(l\right)\\ V_{kk\left(đktc\right)}=\dfrac{100.1,12}{20}=5,6\left(l\right)\\ b,m_{CuO}=0,1.80=8\left(g\right)\\ c,2R+O_2\rightarrow\left(t^o\right)2RO\\ n_R=2.n_{O_2}=2.0,05=0,1\left(mol\right)\\ M_R=\dfrac{2,4}{0,1}=24\left(\dfrac{g}{mol}\right)\\ \Rightarrow R:Magie\left(Mg=24\right)\)
\(a) 4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\\ n_{Al} = \dfrac{5,4}{27} = 0,2(mol)\\ n_{Al_2O_3} = \dfrac{1}{2}n_{Al} = 0,1(mol) \Rightarrow m_{Al_2O_3} = 0,1.102 = 10,2(gam)\\ b) n_{O_2} = \dfrac{3}{4}n_{Al} = 0,15(mol)\\ 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ n_{KMnO_4} = 2n_{O_2} = 0,3(mol) \Rightarrow m_{KMnO_4} = 0,3.158 = 47,4(gam)\)
a, PT: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
Theo PT: \(n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,1\left(mol\right)\)
\(\Rightarrow m_{Al_2O_3}=0,1.102=10,2\left(g\right)\)
b, Theo PT: \(n_{O_2}=\dfrac{3}{4}n_{Al}=0,15\left(mol\right)\)
PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
Theo PT: \(n_{KMnO_4}=2n_{O_2}=0,3\left(mol\right)\)
\(\Rightarrow m_{KMnO_4}=0,2.158=47,4\left(g\right)\)
Bạn tham khảo nhé!
\(n_{Na_2O}=\dfrac{124}{62}=2\left(mol\right)\)
PTHH: 4Na + O2 --to--> 2Na2O
1<----------2
=> mO2 = 1.32 = 32 (g)
a. \(2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2\)
\(n_{KMnO_4}=0,6mol\)
\(\rightarrow n_{O_2}=\frac{1}{2}n_{KMnO_4}=0,3mol\)
\(\rightarrow V_{O_2}=6,72l\)
\(V_{O_2\text{thực}}=\frac{6,72.75}{100}=5,04l\)
b. \(2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2\)
\(n_{O_2}=1,5mol\)
\(\rightarrow n_{KMnO_4}=2n_{O_2}=3mol\)
\(\rightarrow m_{KMnO_4\text{cần}}=\frac{474.100}{80}=592,5g\)
\(n_{KMnO_4}=\dfrac{5,53}{158}=0,035\left(mol\right)\\ 2KMnO_4\rightarrow\left(t^o,xt\right)K_2MnO_4+MnO_2+O_2\\ a,n_{O_2\left(LT\right)}=\dfrac{0,035}{2}=0,0175\left(mol\right)\\ V_{O_2\left(TT\right)}=\left(0,0175.80\%\right).22,4=0,3136\left(l\right)\\ b,2R+O_2\rightarrow\left(t^o\right)2RO\\ n_R=2.0,0175.80\%=0,028\left(mol\right)\\ M_R=\dfrac{0,672}{0,028}=24\left(\dfrac{g}{mol}\right)\\ \Rightarrow R:Magie\left(Mg=24\right)\)