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a) \(n_{SO_2}=\dfrac{6,4}{64}=0,1\left(mol\right)\)
PTHH : S + O2 - to---> SO2
0,1 0,1 0,1 ( mol )
b) \(m_S=0,1.32=3,2\left(g\right)\)
\(V_{O_2}=0,1.22,4=2,24\left(l\right)\)
\(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\\ n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ PTHH:4P+5O_2\underrightarrow{t^o}2P_2O_5\\ LTL:\dfrac{0,2}{4}< \dfrac{0,4}{5}\Rightarrow O_2dư\)
\(n_{O_2\left(pư\right)}=\dfrac{5}{4}n_P=\dfrac{5}{4}.0,2=0,25\left(mol\right)\\ n_{O_2\left(dư\right)}=0,4-0,25=0,15\left(mol\right)\)
\(n_{P_2O_5\left(lt\right)}=\dfrac{1}{2}n_P=\dfrac{1}{2}.0,2=0,1\left(mol\right)\\ m_{P_2O_5\left(lt\right)}=0,1.142=14,2\left(g\right)\\ m_{P_2O_5\left(tt\right)}=0,1.142.80\%=11,36\left(g\right)\)
a) Có chất mới sinh ra
b) Theo ĐLBTKL: mS + mO2 = mSO2
=> mO2 = 6,4 - 3,2 = 3,2 (g)
c) Xét \(d_{O_2/kk}=\dfrac{32}{29}=1,1\)
=> Khí O2 nặng hơn không khí 1,1 lần
PTHH : \(S+O_2\xrightarrow[]{t^o}SO_2\)
\(BTKl:\) \(m_S+m_{O2}=m_{SO2}\)
\(\Rightarrow m_{O2}=m_{SO2}-m_S=6,4-3,2=3,2\left(g\right)\)
\(n_{Al}=\frac{5,4}{27}=0,2\left(mol\right);n_{H_2SO_4}=\frac{19,6}{98}=0,2\left(mol\right)\)
PTPƯ :
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)
2 mol 3 mol
0,2 mol 0,2 mol
0,2/2 > 0,2/3
=> Al dư, bài toán tính theo \(H_2SO_4\)
a. \(n_{Al_2\left(SO_4\right)_3}=\frac{1}{3}n_{H_2SO_{\text{4}}}=\frac{1}{3}.0,2=0,06\left(mol\right)\)
=> \(m_{Al_2\left(SO_4\right)_3}=0,06.342=20,52\left(g\right)\)
b. \(n_{Al\left(TG\right)}=\frac{2}{3}n_{H_2SO_{\text{4}}}=\frac{2}{3}.0,2=0,13\left(mol\right)\)
\(n_{Al\left(dư\right)}=0,2-0,13=0,07\left(mol\right)\)
=> \(m_{Al\left(dư\right)}=0,07.27=1,89\left(g\right)\)
c. \(n_{H_2}=n_{H_2SO_4}=0,2\left(mol\right)\)
Vì hiệu suất đạt 80% nên:
\(n_{H_2}=80\%.0,2=0,16\left(mol\right)\)
\(V_{H_2}=0,16.22,4=3,584\left(l\right)\)
\(n_{Cu}=0,2\left(mol\right)\)
\(Cu+2H_2SO_4\left(đ,đ\right)\underrightarrow{t^0}CuSO_4+SO2+2H_2O\)
\(0,2\) \(0,2\)
\(m_{CuSo_4}=32\left(g\right)\)
Mà thực tế : \(m_{CuSO_4}=28\left(g\right)\)
\(\Rightarrow H=87,5\%\)
a, \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
b, \(n_{O_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)
Theo PT: \(n_{Al_2O_3}=\dfrac{2}{3}n_{O_2}=0,2\left(mol\right)\)
\(\Rightarrow m_{Al_2O_3}=0,2.102=20,4\left(g\right)\)
c, \(H=\dfrac{18,36}{20,4}.100\%=90\%\)
\(n_S=\dfrac{3,2}{32}=0,1\left(mol\right)\)
PT: \(S+O_2\underrightarrow{t^o}SO_2\)
Theo PT: \(n_{SO_2\left(LT\right)}=n_S=0,1\left(mol\right)\)
\(\Rightarrow m_{SO_2\left(LT\right)}=0,1.64=6,4\left(g\right)\)
Mà: H = 80%
\(\Rightarrow m_{SO_2\left(TT\right)}=6,4.80\%=5,12\left(g\right)\)
\(n_S=\dfrac{32}{32}=1mol\\ S+O_2\xrightarrow[]{t^0}SO_2\\ n_{SO_2}=n_S=1mol\\ m_{SO_2\left(lt\right)}=1.64=64g\\ m_{SO_2\left(tt\right)}=64\cdot80:100=51,2g\)