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\(a,PTHH:Fe_3O_4+4H_2\xrightarrow{t^o}3Fe+4H_2O\\ n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol);n_{Fe_3O_4}=\dfrac{46,4}{232}=0,2(mol)\)
Vì \(\dfrac{n_{H_2}}{4}<\dfrac{n_{Fe_3O_4}}{1}\) nên \(Fe_3O_4\) dư
\(n_{Fe_3O_4(dư)}=0,2-\dfrac{0,3}{4}=0,125(mol)\\ \Rightarrow m_{Fe_3O_4(dư)}=0,125.232=29(g)\\ b,n_{Fe}=\dfrac{3}{4}n_{H_2}=0,225(mol)\\ \Rightarrow m_{Fe}=0,225.56=12,6(g)\)
Cảm ơn bạn @anayuiky đã nhắc lỗi sai. Mình sửa lại ý c):
PTHH: \(2KMnO_4\rightarrow^{t^o}K_2MnO_4+MnO_2+O_2\uparrow\)
Theo phương trình \(n_{KMnO_4}=n_{O_2}.2=0,25.2=0,5mol\)
\(\rightarrow m_{KMnO_4}=0,5.\left(39+55+16.4\right)=79g\)
a. \(n_{H_2}=\frac{V}{22,4}=\frac{11,2}{22,4}=0,5mol\)
\(n_{O_2}=\frac{V}{22,4}=\frac{10,08}{22,4}=0,45mol\)
PTHH: \(2H_2+O_2\rightarrow^{t^o}2H_2O\)
Ban đầu: 0,5 0,45 mol
Trong pứng: 0,5 0,25 0,5 mol
Sau pứng: 0 0,2 0,5 mol
\(\rightarrow M_{O_2\left(dư\right)}=n.M=0,2.32=6,4g\)
b. Theo phương trình \(n_{H_2O}=n_{H_2}=0,5mol\)
\(\rightarrow m_{H_2O}=n.M=0,5.18=9g\)
c. PTHH: \(2KMnO_4\rightarrow^{t^o}K_2MnO_4+MnO_2+O_2\uparrow\)
0,9 0,45 mol
\(\rightarrow n_{KMnO_4}=\frac{2}{1}n_{O_2}=\frac{0,45.2}{1}=0,9mol\)
\(\rightarrow m_{KMnO_4}=n.M=0,9.158=142,2g\)
a, Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PT: \(2H_2+O_2\underrightarrow{t^o}2H_2O\)
Xét tỉ lệ: \(\dfrac{0,3}{2}< \dfrac{0,2}{1}\), ta được O2 dư.
Theo PT: \(n_{O_2\left(pư\right)}=\dfrac{1}{2}n_{H_2}=0,15\left(mol\right)\)
\(\Rightarrow n_{O_2\left(dư\right)}=0,05\left(mol\right)\Rightarrow m_{O_2\left(dư\right)}=0,05.32=1,6\left(g\right)\)
b, \(n_{H_2O}=n_{H_2}=0,3\left(mol\right)\)
\(\Rightarrow m_{H_2O}=0,3.18=5,4\left(g\right)\)
c, PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
_______0,3_______________________0,15 (mol)
\(\Rightarrow m_{KMnO_4}=0,3.158=47,4\left(g\right)\)
Bạn tham khảo nhé!
a) \(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
\(n_{O_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\)
PTHH: 2H2 + O2 --to--> 2H2O
Xét tỉ lệ: \(\dfrac{0,5}{2}< \dfrac{0,45}{1}\) => H2 hết, O2 dư
PTHH: 2H2 + O2 --to--> 2H2O
0,5-->0,25----->0,5
=> \(m_{O_2\left(dư\right)}=\left(0,45-0,25\right).32=6,4\left(g\right)\)
b) \(m_{H_2O}=0,5.18=9\left(g\right)\)
c)
PTHH: 2KMnO4 --to--> K2MnO4 + MnO2 + O2
0,5<-----------------------------------0,25
=> \(m_{KMnO_4}=0,5.158=79\left(g\right)\)
\(a.n_{CuO}=\dfrac{40}{80}=0,5\left(mol\right)\\ n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ PTHH:CuO+H_2\underrightarrow{t^o}Cu+H_2O\\ Vì:\dfrac{0,15}{1}< \dfrac{0,5}{1}\\ \rightarrow CuOdư\\ n_{CuO\left(p.ứ\right)}=n_{Cu}=n_{H_2}=0,15\left(mol\right)\\ \rightarrow n_{CuO\left(dư\right)}=0,5-0,15=0,35\left(mol\right)\\ m_{CuO\left(DƯ\right)}=0,35.80=28\left(g\right)\\ b.m_{Cu}=0,35.64=22,4\left(g\right)\\ c.m_{hh_{rắn}}=m_{Cu}+m_{CuO\left(dư\right)}=22,4+28=50,4\left(g\right)\)
\(n_{Fe}=\dfrac{12.6}{56}=0.225\left(mol\right)\)
\(n_{O_2}=\dfrac{4.2}{22.4}=0.1875\left(mol\right)\)
\(3Fe+2O_2\underrightarrow{^{^{t^0}}}Fe_3O_4\)
\(3.........2\)
\(0.225......0.1875\)
Lập tỉ lệ : \(\dfrac{0.225}{3}< \dfrac{0.1875}{2}\Rightarrow O_2dư\)
\(m_{O_2\left(dư\right)}=\left(0.1875-0.225\cdot\dfrac{2}{3}\right)\cdot32=1.2\left(g\right)\)
\(m_{Fe_3O_4}=\dfrac{0.225}{3}\cdot232=17.4\left(g\right)\)
\(nAl=\dfrac{8,1}{27}=0,3\left(mol\right)\)
\(nHCl=\dfrac{21,9}{36,5}=0,6\left(mol\right)\)
PTHH:
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
2 6 2 3 (mol)
0,2 0,6 0,2 0,3 (mol)
LTL : 0,3 / 2 > 0,6/6
=> Al dư sau pứ , HCl đủ vs pứ
\(mAl_{\left(dư\right)}=\left(0,3-0,2\right).27=2,7\left(g\right)\)
\(mAlCl_3=0,2.98=19,6\left(g\right)\)
\(H_2+CuO\rightarrow Cu+H_2O\)
1 1 1 1 (mol)
0,3 0,3 0,3 0,3 (mol)
=> \(mCu=0,3.64=19,2\left(g\right)\)
\(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\\
n_{HCl}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\\
pthh:2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
\(LTL:\dfrac{0,3}{2}>\dfrac{0,6}{6}\)
=> Al dư HCl hết
theo pthh : \(n_{Al\left(p\text{ư}\right)}=\dfrac{1}{3}n_{HCl}=0,2\left(mol\right)\\ m_{Al\left(d\right)}=\left(0,3-0,2\right).27=2,7\left(g\right)\)
theo pthh : \(n_{AlCl_3}=\dfrac{1}{3}n_{HCl}=0,1\left(mol\right)\\
m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\)
theo pthh : \(n_{H_2}=\dfrac{1}{2}n_{HCl}=0,3\left(mol\right)\)
pthh: \(CuO+H_2\underrightarrow{t^o}H_2O+Cu\)
0,3 0,3
\(m_{Cu}=0,3.64=19,2\)
Câu 1
\(n_{Fe}=\dfrac{16,8}{56}=0,3mol\\ n_{O_2}=\dfrac{2,479}{24,79}=0,1mol\\ 3Fe+2O_2\xrightarrow[t^0]{}Fe_3O_4\\ \Rightarrow\dfrac{0,3}{3}>\dfrac{0,1}{2}\Rightarrow Fe.dư\\ 3Fe+2O_2\xrightarrow[t^0]{}Fe_3O_4\)
0,15 0,1 0,05
\(m_{Fe.dư}=16,8-0,15.56=8,4g\\ b.m_{Fe_3O_4}=0,05.232=11,6g\)
\(n_{H_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\\ n_{O_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\)
PTHH: 2H2 + O2 --to--> 2H2O
LTL: \(\dfrac{1,5}{2}< 1,5\rightarrow O_2\) dư
Theo pt: \(\left\{{}\begin{matrix}n_{O_2\left(pư\right)}=\dfrac{1}{2}n_{H_2}=\dfrac{1}{2}.1,5=0,75\left(mol\right)\\n_{H_2O}=n_{H_2}=1,5\left(mol\right)\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}m_{O_2\left(dư\right)}=\left(1,5-0,75\right).32=24\left(g\right)\\V_{O_2}\left(1,5-0,75\right).22,4=16,8\left(l\right)\\m_{H_2O}=1,5.18=27\left(g\right)\end{matrix}\right.\)
\(n_{H_2}=n_{O_2}=\dfrac{33,6}{22,4}=1,5\left(MOL\right)\)
pthh: \(2H_2+O_2\underrightarrow{t^O}2H_2O\)
LTL : \(\dfrac{1,5}{2}< \dfrac{1,5}{1}\)
=> O2 dư , H2 hết
theo pthh: nH2O = nH2 = 1,5 (mol)
=> \(m_{H_2O}=1,5.18=27\left(g\right)\)
Cíu pls