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KCl + O2
$n_{KClO_3} = \dfrac{12,25}{122,5} = 0,1(mol)$
$2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2$
$n_{O_2} = 1,5n_{KClO_3} = 0,15(mol)$
$n_P = 0,5(mol)$
$4P + 5O_2 \xrightarrow{t^o} 2P_2O_5$
$n_P : 4 > n_{O_2} : 5$ nên P dư
$n_{P_2O_5} = \dfrac{2}{5}n_{O_2} = 0,06(mol)$
$m_{P_2O_5} = 0,06.142 = 8,52(gam)$
a, PT: \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
Ta có: \(n_{KClO_3}=\dfrac{24,5}{122,5}=0,2\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{3}{2}n_{KClO_3}=0,3\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,3.24,79=7,437\left(g\right)\)
b, PT: \(2Cu+O_2\underrightarrow{t^o}2CuO\)
Ta có: \(n_{Cu}=\dfrac{32}{64}=0,5\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,5}{2}< \dfrac{0,3}{1}\), ta được O2 dư.
Theo PT: \(\left\{{}\begin{matrix}n_{O_2\left(pư\right)}=\dfrac{1}{2}n_{Cu}=0,25\left(mol\right)\\n_{CuO}=n_{Cu}=0,5\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{O_2\left(dư\right)}=0,3-0,25=0,05\left(mol\right)\)
\(\Rightarrow m_{O_2\left(dư\right)}=0,05.32=1,6\left(g\right)\)
\(m_{CuO}=0,5.80=40\left(g\right)\)
\(n_{Al}=\dfrac{1,728}{27}=0,064\left(mol\right)\)
PTHH: 4Al + 3O2 --to--> 2Al2O3
____0,064->0,048
=> mO2 = 0,048.32 = 1,536 (g)
\(m_B=\dfrac{0,894.100}{8,127}=11\left(g\right)\)
Theo ĐLBTKL: mA = mB + mO2
=> mA = 11 + 1,536 = 12,536 (g)
\(a,PTHH:2KClO_3\xrightarrow{t^o}2KCl+3O_2\\ b,n_{KClO_3(thực tế)}=\dfrac{36,75}{122,5}=0,3(mol)\\ n_{O_2(phản ứng)}=\dfrac{6,72}{22,4}=0,3(mol)\\ \Rightarrow n_{KClO_3(phản ứng)}=\dfrac{2}{3}n_{O_2}=0,2(mol)\\ \Rightarrow H\%=\dfrac{0,2}{0,3}.100\%=66,67\%\\ c,n_{KCl}=n_{KClO_3(phản ứng)}=0,2(mol)\\ \Rightarrow m_{KCl}=0,2.74,5=14,9(g)\)
`#3107.101107`
1.
a.
Ta có:
\(\text{n}_{\text{KClO}_3}=\dfrac{\text{m}_{\text{KClO}_3}}{\text{M}_{\text{KClO}_3}}=\dfrac{122,5}{122,5}=1\text{ (mol)}\)
PTPỨ: \(\text{2KClO}_3\text{ }\)\(\underrightarrow{\text{ }t^0}\) \(\text{2KCl}+3\text{O}_2\)
Ta có: `2` mol \(\text{KClO}_3\) thu được `3` mol \(\text{O}_2\)
`=>` `1` mol \(\text{KClO}_3\) thu được `1,5` mol \(\text{O}_2\)
b.
\(\text{V}_{\text{O}_2}=\text{n}_{\text{O}_2}\cdot24,79=1,5\cdot24,79=37,185\left(l\right)\)
TTĐ:
\(m_{KClO_3}=122,5\left(g\right)\)
______________
a) PTHH?
b) \(V_{O_2}=?\left(l\right)\)
Giải
\(n_{KClO_3}=\dfrac{m}{M}=\dfrac{122,5}{122,5}=1\left(mol\right)\)
\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\uparrow\)
1-> 1 : 1,5(mol)
\(V_{O_2}=n.22,4=1,5.22,4=33,6\left(l\right)\)
$n_{KClO_3} = \dfrac{12,25}{122,5} = 0,1(mol)$
$2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2$
Theo PTHH : $n_{O_2} = \dfrac{3}{2}n_{KClO_3} = 0,15(mol)$
$n_P = \dfrac{15,5}{31} = 0,5(mol)$
$4P + 5O_2 \xrightarrow{t^o} 2P_2O_5$
Ta thấy : $n_P : 4 > n_{O_2} : 5$ nên P dư
$n_{P_2O_5} = \dfrac{2}{5}n_{O_2} = 0,06(mol)$
$m_{P_2O_5} = 0,06.142 = 8,52(gam)$
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