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\(n_P=\dfrac{12,4}{31}=0,4\left(mol\right)\\ a,PTHH:4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\\ b,n_{O_2}=\dfrac{5}{4}.0,4=0,5\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,5.22,4=11,2\left(l\right)\\ c,n_{P_2O_5}=\dfrac{2}{4}.0,4=0,2\left(mol\right)\\ m_{P_2O_5}=142.0,2=28,4\left(g\right)\)
Bài 2:
\(n_{Al}=\dfrac{16,2}{27}=0,6\left(mol\right)\\ n_{O_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\\ a,4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\\ Vì:\dfrac{0,6}{4}< \dfrac{0,6}{3}\Rightarrow O_2dư\\ n_{O_2\left(dư\right)}=0,6-\dfrac{3}{4}.0,6=0,15\left(mol\right)\\ \Rightarrow m_{O_2\left(dư\right)}=0,15.32=4,8\left(g\right)\\ c,n_{Al_2O_3}=\dfrac{2}{4}.n_{Al}=\dfrac{2}{4}.0,6=0,3\left(mol\right)\\ \Rightarrow m_{Al_2O_3}=102.0,3=30,6\left(g\right)\)
Bài 1.
\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1mol\)
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
0,1 0,6 0,2
\(m_{HCl}=0,6\cdot36,5=21,9g\)
\(m_{FeCl_3}=0,2\cdot162,5=32,5g\)
\(n_{O_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\)
PTHH: 4Al + 3O2 --to--> 2Al2O3
2<---1,5---------->1
=> \(m_{Al_2O_3}=1.102=102\left(g\right)\)
\(m_{Al}=2.27=54\left(g\right)\)
\(a,PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ \Rightarrow n_{Fe}=n_{FeCl_2}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ \Rightarrow m_{FeCl_2}=0,2\cdot127=25,4\left(g\right)\\ b,n_{H_2}=n_{Fe}=0,2\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}m_{H_2}=0,2\cdot2=0,4\left(g\right)\\V_{H_2\left(đktc\right)}=0,2\cdot22,4=4,48\left(l\right)\end{matrix}\right.\)
\(c,PTHH:2H_2+O_2\rightarrow^{t^0}2H_2O\\ \Rightarrow n_{O_2}=\dfrac{1}{2}n_{H_2}=0,1\left(mol\right)\\ \Rightarrow V_{O_2\left(đktc\right)}=0,1\cdot22,4=2,24\left(l\right)\)
a.b.c.\(n_{Mg}=\dfrac{m_{Mg}}{M_{Mg}}\dfrac{4,8}{24}=0,2mol\)
\(2Mg+O_2\rightarrow\left(t^o\right)2MgO\)
0,2 0,1 0,2 ( mol )
\(V_{O_2}=n_{O_2}.22,4=0,1.22,4=2,24l\)
\(m_{MgO}=n_{MgO}.M_{MgO}=0,2.40=8g\)
d. Sửa đề: tính khối lượng KMnO4
\(2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\)
0,2 0,1 ( mol )
\(m_{KMnO_4}=n_{KMnO_4}.M_{KMnO_4}=0,2.158=31,6g\)
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
\(PTHH:2Mg+O_2-^{t^O}>2MgO\)
tỉ lệ: 2 : 1 : 2
n(mol) 0,2---->0,1----->0,2
\(m_{MgO}=n\cdot M=0,2\cdot\left(24+16\right)=8\left(g\right)\)