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a) $n_P = \dfrac{6,2}{31} = 0,2(mol) ; n_{O_2} = \dfrac{6,4}{32} = 0,2(mol)$
$4P + 5O_2 \xrightarrow{t^o} 2P_2O_5$
$n_{P\ pư} = \dfrac{4}{5}n_{O_2} = 0,16(mol)$
$\Rightarrow m_{P\ dư} = 6,2 -0,16.31 = 1,24(gam)$
b) Sản phẩm là $P_2O_5$
$n_{P_2O_5} = \dfrac{2}{5}n_{O_2} = 0,08(mol)$
$m_{P_2O_5} = 0,08.142 = 11,36(gam)$
PTHH: \(2Zn+O_2\underrightarrow{t^o}2ZnO\)
Ta có: \(\left\{{}\begin{matrix}n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\n_{O_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,1}{2}< \dfrac{0,1}{1}\) \(\Rightarrow\) Oxi còn dư, Zn p/ứ hết
\(\Rightarrow\left\{{}\begin{matrix}n_{O_2\left(dư\right)}=0,05\left(mol\right)\\n_{ZnO}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{O_2\left(dư\right)}=0,05\cdot32=1,6\left(g\right)\\m_{ZnO}=0,1\cdot81=8,1\left(g\right)\end{matrix}\right.\)
a, Theo giả thiết ta có: \(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)
\(4P+5O_2--t^o->2P_2O_5\)
Ta có: \(n_{O_2}=\dfrac{5}{4}.n_P=0,125\left(mol\right)\Rightarrow V_{O_2\left(đktc\right)}=0,125.22,4=2,8\left(l\right)\)
b, Theo giả thiết ta có: \(n_{CH_4}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
\(CH_4+2O_2--t^o->CO_2+2H_2O\)
Ta có: \(n_{O_2}=2.n_{CH_4}=0,1\left(mol\right)\Rightarrow V_{O_2\left(đktc\right)}=2,24\left(l\right)\)
PTHH : \(4Fe+3O_2\left(t^o\right)-->2Fe_2O_3\)
\(n_{Fe}=\dfrac{m}{M}=\dfrac{16.8}{56}=0.3\left(mol\right)\)
\(n_{O_2}=\dfrac{V}{22.4}=\dfrac{17.92}{22.4}=0.8\left(mol\right)\)
Có \(n_{Fe}< n_{O_2}\) (0.3 < 0.8) => O2 dư , Fe hết
\(n_{O_2\left(dư\right)}=n_{O_2\left(PƯ\right)}-n_{Fe}=0.8-0.3=0.5\left(mol\right)\)
=> \(m_{O_2\left(dư\right)}=n_{O_2\left(dư\right)}.M=16\left(g\right)\)
Sản phẩm thu đc lak Fe2O3
Từ PTHH => \(\dfrac{1}{2}n_{Fe}=n_{Fe_2O_3}=0.15\left(mol\right)\)
=> \(m_{Fe_2O_3}=n.M=0,15.\left(56.2+16.3\right)=24\left(g\right)\)
mik c.ơn trước
a)
$2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2$
b) $n_{KMnO_4} = \dfrac{79}{158} = 0,5(mol)$
Theo PTHH : $n_{O_2} = \dfrac{1}{2}n_{KMnO_4} = 0,25(mol)$
$\Rightarrow V_{O_2} = 0,25.22,4 = 5,6(lít)$
c) $n_P = \dfrac{3,1}{31} = 0,1(mol)$
$4P + 5O_2 \xrightarrow{t^o} 2P_2O_5$
Ta thấy : $n_P : 4 < n_{O_2} :5$ nên $O_2$ dư
$n_{P_2O_5} = \dfrac{1}{2}n_P = 0,05(mol)$
$m_{P_2O_5} = 0,05.142 = 7,1(gam)$
\(n_{H_2}=\dfrac{V}{24,79}=\dfrac{11,2}{24,79}\approx0,45\left(mol\right)\)
a) \(PTHH:2H_2+O_2\underrightarrow{t^o}2H_2O\)
2 1 2
0,45 0,225 0,45
b) \(m_{O_2}=n.M=0,225.\left(16.2\right)=7,2\left(g\right)\\ V_{O_2}=n.24,79=0,225.24,79=5,57775\left(l\right)\)
c) \(PTHH:2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2\)
2 1 1 1
0,45 0,225 0,225 0,225
\(m_{KMnO_4}=n.M=0,45.\left(39+55+16.4\right)=71,1\left(g\right).\)
a, \(2H_2+O_2\underrightarrow{t^o}2H_2O\)
b, Ta có: \(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{H_2}=0,25\left(mol\right)\)
\(\Rightarrow m_{O_2}=0,25.32=8\left(g\right)\)
\(V_{O_2}=0,25.22,4=5,6\left(l\right)\)
c, \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
Theo PT: \(n_{KMnO_4}=2n_{O_2}=0,5\left(mol\right)\Rightarrow m_{KMnO_4}=0,5.158=79\left(g\right)\)
\(n_{CO_2}=\dfrac{8.96}{22.4}=0.4\left(mol\right)\)
\(n_{H_2O}=\dfrac{9}{18}=0.5\left(mol\right)\)
\(n_{O_2}=\dfrac{14.56}{22.4}=0.65\left(mol\right)\)
\(BTKL:\)
\(m_A=0.4\cdot44+9-0.65\cdot32=5.8\left(g\right)\)
\(m_O=5.8-0.4\cdot12-0.5\cdot2=0\)
\(n_A=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)
Số nguyên tử C : \(\dfrac{0.4}{0.1}=4\)
Số nguyên tử H : \(\dfrac{0.5\cdot2}{0.1}=10\)
\(CT:C_4H_{10}\)
\(a,PTHH:4K+O_2\underrightarrow{t^o}2K_2O\\ b,n_{O_2}=\dfrac{V_{\left(đktc\right)}}{22,4}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ Theo.PTHH:n_K=4n_{O_2}=4.0,1=0,4\left(mol\right)\\ \Rightarrow m_K=n.M=0,4.39=15,6\left(g\right)\\ c,Theo.PTHH:n_{K_2O}=2n_{O_2}=2.0,1=0,2\left(mol\right)\\ \Rightarrow m_{K_2O}=n.M=0,2.94=18,8\left(g\right)\)
$2Zn + O_2 \xrightarrow{t^o} 2ZnO$
$n_{O_2} = \dfrac{2,24}{22,4} = 0,1(mol)$
$n_{Zn} = 2n_{O_2} = 0,2(mol)$
$\Rightarrow a = 0,2.65 = 13(gam)$
Cách 1 : $n_{ZnO} = n_{Zn} = 0,2(mol)$
$m_{ZnO} = 0,2.81 = 16,2(gam)$
Cách 2 : $m_{ZnO} = m_{Zn} + m_{O_2} = 13 + 0,1.32 = 16,2(gam)$