Bài 8:
\(n_P=\dfrac{m_P}{M_P}=\dfrac{6,2}{31}=0,2\left(mol\right)\\ 4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\\ a,n_{O_2}=\dfrac{5}{4}.n_P=\dfrac{5}{4}.0,2=0,25\left(mol\right)\\ \Rightarrow V_{O_2\left(đktc\right)}=n_{O_2}.22,4=0,25.22,4=5,6\left(l\right)\\ b,n_{P_2O_5}=\dfrac{2}{4}.n_P=\dfrac{2}{4}.0,2=0,1\left(mol\right)\\ \Rightarrow m_{P_2O_5}=M_{P_2O_5}.n_{P_2O_5}=142.0,1=14,2\left(g\right)\)
Bài 7:
\(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ PTHH:4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\\ a,n_{O_2}=\dfrac{3}{4}.n_{Al}=\dfrac{3}{4}.0,2=0,15\left(mol\right)\\ \Rightarrow V_{O_2\left(đktc\right)}=n_{O_2}.22,4=0,15.22,4=3,36\left(l\right)\\ b,n_{Al_2O_3}=\dfrac{n_{Al}}{2}=\dfrac{0,2}{2}=0,1\left(mol\right)\\ \Rightarrow m_{Al_2O_3}=n_{Al_2O_3}.M_{Al_2O_3}=0,1.102=10,2\left(g\right)\)
Stop
Em ơi
Đây có 7 bài, em làm được bài nào rồi? Chứ chưa làm được bài nào thật hả?
Bài 6:
\(n_{Fe}=\dfrac{m_{Fe}}{M_{Fe}}=\dfrac{22,4}{56}=0,4\left(mol\right)\\ a,Fe+2HCl\rightarrow FeCl_2+H_2\\ b,n_{H_2}=n_{Fe}=0,4\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=n_{H_2}.22,4=0,4.22,4=8,96\left(l\right)\\ c,n_{HCl}=2.n_{Fe}=2.0,4=0,8\left(mol\right)\\ \Rightarrow m_{HCl}=n_{HCl}.M_{HCl}=0,8.36,5=29,2\left(g\right)\)
Bài 5:
\(n_{CuO}=\dfrac{m_{CuO}}{M_{CuO}}=\dfrac{20}{80}=0,25\left(mol\right)\\a,PTHH: CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\\ b,n_{H_2}=n_{CuO}=0,25\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=n_{H_2}.22,4=0,25.22,4=5,6\left(l\right)\)
Bài 4:
\(n_{Fe_2O_3}=\dfrac{m_{Fe_2O_3}}{M_{Fe_2O_3}}=\dfrac{16}{160}=0,1\left(mol\right)\\ a,PTHH:4Fe+3O_2\left(thiếu\right)\rightarrow\left(t^o\right)2Fe_2O_3\\b,n_{Fe}=\dfrac{4}{2}.n_{Fe_2O_3}=\dfrac{4}{2}.0,1=0,2\left(mol\right)\\ \Rightarrow m_{Fe}=M_{Fe}.n_{Fe}=56.0,2=11,2\left(g\right)\\ c,n_{O_2}=\dfrac{3}{2}.n_{Fe_2O_3}=\dfrac{3}{2}.0,1=0,15\left(mol\right)\\ \Rightarrow V_{O_2\left(đktc\right)}=n_{O_2}.22,4=0,15.22,4=3,36\left(l\right)\)
Bài 3:
\(n_{CaCO_3}=\dfrac{m_{CaCO_3}}{M_{CaCO_3}}=\dfrac{10}{100}=0,1\left(mol\right)\\ PTHH:CaCO_3\rightarrow\left(t^o\right)CaO+CO_2\uparrow\\ Ta.có:n_{CaO}=n_{CO_2}=n_{CaCO_3}=0,1\left(mol\right)\\ a,V_{CO_2\left(đktc\right)}=n_{CO_2}.22,4=0,1.22,4=2,24\left(l\right)\\ b,m_{CaO}=n_{CaO}.M_{CaO}=0,1.56=5,6\left(g\right)\)
Bài 2:
\(a,Đặt.CTTQ:S_xO_y\left(x,y:nguyên,dương\right)\\ Theo.đề.bài,ta.có:\\ x=\dfrac{40\%.80}{32}=1;y=\dfrac{60\%.80}{16}=3\\ \Rightarrow CTHH:SO_3\\ b,Đặt.CTTQ:K_aN_bO_c\\ Theo.đề.bài,ta.có:\\ a=\dfrac{101.38,61\%}{39}=1;b=\dfrac{13,86\%.101}{14}=1\\ c=\dfrac{\left(100\%-38,61\%-13,86\%\right).101}{16}=3\\ \Rightarrow CTHH:KNO_3\)
