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\(a.n_{CuO}=\dfrac{40}{80}=0,5\left(mol\right)\\ n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ PTHH:CuO+H_2\underrightarrow{t^o}Cu+H_2O\\ Vì:\dfrac{0,15}{1}< \dfrac{0,5}{1}\\ \rightarrow CuOdư\\ n_{CuO\left(p.ứ\right)}=n_{Cu}=n_{H_2}=0,15\left(mol\right)\\ \rightarrow n_{CuO\left(dư\right)}=0,5-0,15=0,35\left(mol\right)\\ m_{CuO\left(DƯ\right)}=0,35.80=28\left(g\right)\\ b.m_{Cu}=0,35.64=22,4\left(g\right)\\ c.m_{hh_{rắn}}=m_{Cu}+m_{CuO\left(dư\right)}=22,4+28=50,4\left(g\right)\)
\(a,PTHH:Fe_3O_4+4H_2\xrightarrow{t^o}3Fe+4H_2O\\ n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol);n_{Fe_3O_4}=\dfrac{46,4}{232}=0,2(mol)\)
Vì \(\dfrac{n_{H_2}}{4}<\dfrac{n_{Fe_3O_4}}{1}\) nên \(Fe_3O_4\) dư
\(n_{Fe_3O_4(dư)}=0,2-\dfrac{0,3}{4}=0,125(mol)\\ \Rightarrow m_{Fe_3O_4(dư)}=0,125.232=29(g)\\ b,n_{Fe}=\dfrac{3}{4}n_{H_2}=0,225(mol)\\ \Rightarrow m_{Fe}=0,225.56=12,6(g)\)
\(n_{CuO}=2a\left(mol\right)\Rightarrow n_{Fe_2O_3}=a\left(mol\right)\)
\(m_X=80\cdot2a+160a=80\left(g\right)\)
\(\Rightarrow a=0.25\left(mol\right)\)
\(CuO+H_2\underrightarrow{^{^{t^0}}}Cu+H_2O\)
\(Fe_2O_3+3H_2\underrightarrow{^{^{t^0}}}2Fe+3H_2O\)
\(n_{H_2}=0.5+0.25\cdot3=1.25\left(mol\right)\)
\(V_{H_2}=1.25\cdot22.4=28\left(l\right)\)
\(m_{cr}=0.5\cdot64+0.5\cdot56=60\left(g\right)\)
\(n_{CuO}=4a\left(mol\right)\Rightarrow n_{FeO}=a\left(mol\right)\)
\(m_X=80\cdot4a+72a=19.6\left(g\right)\)
\(\Rightarrow a=0.05\)
\(CuO+H_2\underrightarrow{^{^{t^0}}}Cu+H_2O\)
\(FeO+H_2\underrightarrow{^{^{t^0}}}Fe+H_2O\)
\(m_{cr}=0.2\cdot64+0.05\cdot56=15.6\left(g\right)\)
\(V_{H_2}=\left(0.05\cdot4+0.05\right)\cdot22.4=5.6\left(l\right)\)
\( CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\\ n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\\ n_{CuO}=\dfrac{48}{80}=0,6\left(mol\right)\\ Vì:\dfrac{0,5}{1}< \dfrac{0,6}{1}\Rightarrow CuO\left(dư\right)\Rightarrow Tính.theo.n_{H_2}\\ Đặt:a=n_{CuO\left(p.ứ\right)}\\ m_{rắn}=41,6\left(g\right)\\ \Leftrightarrow64a+80.\left(0,6-a\right)=41,6\\ \Leftrightarrow a=0,4\left(mol\right)\\ n_{CuO\left(LT\right)}=n_{H_2}=0,5\left(mol\right)\\ \Rightarrow H=\dfrac{n_{CuO\left(TT\right)}}{n_{CuO\left(LT\right)}}.100\%=\dfrac{0,4}{0,5}.100=80\%\)
Thể tích H2 phản ứng: 11,2 (lít) (đề bài)
\( \%m_{CuO\left(p.ứ\right)}=\dfrac{0,4}{0,6}.100\%=66,667\%\) (Do số mol tỉ lệ thuận với khối lượng)
a, PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
Ta có: \(n_{CuO}=\dfrac{3,2}{80}=0,04\left(mol\right)\)
Theo PT: \(n_{Cu}=n_{H_2O}=n_{CuO}=0,04\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Cu}=0,04.64=2,56\left(g\right)\\m_{H_2O}=0,04.18=0,72\left(g\right)\end{matrix}\right.\)
b, PT: \(Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\)
Ta có: \(n_{Fe_3O_4}=\dfrac{10,8}{232}=\dfrac{27}{580}\left(mol\right)\)
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{\dfrac{27}{580}}{1}< \dfrac{0,2}{4}\), ta được H2 dư.
Theo PT: \(\left\{{}\begin{matrix}n_{H_2\left(pư\right)}=n_{H_2O}=4n_{Fe_3O_4}=\dfrac{27}{145}\left(mol\right)\\n_{Fe}=3n_{Fe_3O_4}=\dfrac{81}{580}\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{H_2\left(dư\right)}=0,2-\dfrac{27}{145}=\dfrac{2}{145}\left(mol\right)\)
\(\Rightarrow m_{H_2\left(dư\right)}=\dfrac{2}{145}.2\approx0,0276\left(g\right)\)
\(m_{H_2O}=\dfrac{27}{145}.18\approx3,35\left(g\right)\)
\(m_{Fe}=\dfrac{81}{580}.56\approx7,82\left(g\right)\)
Bạn tham khảo nhé!
a) 4P + 5O2 --to--> 2P2O5
b) \(n_{O_2}=\dfrac{16}{32}=0,5\left(mol\right)\)
PTHH: 4P + 5O2 --to--> 2P2O5
0,4<--0,5------->0,2
=> mP2O5 = 0,2.142 = 28,4 (g)
c_ mP = 0,4.31 = 12,4 (g)
\(n_{O_2}=\dfrac{16}{32}=0,5\left(mol\right)\)
PTHH: 4P + 5O2 --to--> 2P2O5
0,4 0,5 0,2
\(\rightarrow\left\{{}\begin{matrix}m_{P_2O_5}=0,2.142=28,4\left(g\right)\\m_P=0,4.31=12,4\left(g\right)\end{matrix}\right.\)
\(n_{H_2}=\dfrac{5.6}{22.4}=0.25\left(mol\right)\)
\(n_{CuO}=\dfrac{7.2}{80}=0.09\left(mol\right)\)
\(CuO+H_2\underrightarrow{^{^{t^0}}}Cu+H_2O\)
Lập tỉ lệ :
\(\dfrac{0.25}{1}>\dfrac{0.09}{1}\rightarrow H_2dư\)
\(n_{Cu}=n_{CuO}=0.09\left(mol\right)\)
\(m=0.09\cdot64=5.76\left(g\right)\)
\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ n_{CuO}=\dfrac{7,2}{80}=0,09\left(mol\right)\\ PTHH:CuO+H_2\underrightarrow{to}Cu+H_2O\\ Vì:\dfrac{0,25}{1}>\dfrac{0,09}{1}\\ \rightarrow H_2dư\\ \rightarrow n_{Cu}=n_{CuO}=0,09\left(mol\right)\\ m_{Cu}=0,09.64=5,76\left(g\right)\)
a) \(CuO+H_2-^{t^o}\rightarrow Cu+H_2O\)
\(n_{CuO\left(bđ\right)}=\dfrac{16}{80}=0,2\left(mol\right)\)
\(\Rightarrow n_{CuO\left(pứ\right)}=0,2.80\%=0,16\left(mol\right)\)
\(n_{H_2O}=n_{CuO}=0,16\left(mol\right)\)
=> \(m_{H_2O}=0,16.18=2,88\left(g\right)\)
b) \(n_{H_2}=0,15\left(mol\right)\)
Lập tỉ lệ : \(\dfrac{0,2}{1}>\dfrac{0,15}{1}\Rightarrow\)Sau phản ứng CuO dư
Chất rắn sau phản ứng là Cu, CuO dư
\(m_{cr}=0,15.64+\left(0,2-0,15\right).80=13,6\left(g\right)\)
c) Gọi x là số mol CuO phản ứng
\(m_{cr}=\left(0,2-x\right).80+64x=13,28\)
=> x=0,17 (mol)
\(H=\dfrac{0,17}{0,2}.100=85\%\)
H2+CuO->Cu+H2O
0,2---0,2----0,2---0,2
n H2=0,3 mol
n CuO=0,2 mol
=>H2 dư
m H2=0,1.2=2g
m Cu=0,2.64=12,8g