PT: \(2K+2H_2O\rightarrow2KOH+H_2\)
\(Ba+2H_2O\rightarrow Ba\left(OH\right)_2+H_2\)
a, Giả sử: \(\left\{{}\begin{matrix}n_K=x\left(mol\right)\\n_{Ba}=y\left(mol\right)\end{matrix}\right.\)
⇒ 39x + 137y = 21,5 (1)
Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Theo PT: \(n_{H_2}=\dfrac{1}{2}n_K+n_{Ba}=\dfrac{1}{2}x+y\left(mol\right)\)
\(\Rightarrow\dfrac{1}{2}x+y=0,2\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,2\left(mol\right)\\y=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\%m_K=\dfrac{0,2.39}{21,5}.100\%\approx36,28\%\)
b, Theo PT: \(\left\{{}\begin{matrix}n_{KOH}=n_K=0,2\left(mol\right)\\n_{Ba\left(OH\right)_2}=n_{Ba}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}C_{M_{KOH}}=\dfrac{0,2}{0,2}=1M\\C_{M_{Ba\left(OH\right)_2}}=\dfrac{0,1}{0,2}=0,5M\end{matrix}\right.\)
c, Ta có: \(n_{OH}=n_{KOH}+2n_{Ba\left(OH\right)_2}=0,4\left(mol\right)\)
Bạn tham khảo nhé!
