PTHH: \(K_2O+H_2O\rightarrow2KOH\)
Ta có: \(n_{K_2O}=\dfrac{x}{94}\left(mol\right)\) \(\Rightarrow n_{KOH\left(thêm\right)}=\dfrac{x}{47}\left(mol\right)\) \(\Rightarrow m_{KOH\left(thêm\right)}=\dfrac{56x}{47}\left(g\right)\)
Theo đề bài: \(\Sigma m_{KOH}=120\cdot51\%=61,2\left(g\right)\)
Ta lập được hệ phương trình: \(\left\{{}\begin{matrix}\dfrac{56x}{47}+12\%\cdot y=61,2\\x+y=120\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\approx43,68\\y\approx76,32\end{matrix}\right.\)
bạn xem lại số liệu nhé chứ số như kia xấu quá :((
Gọi \(n_{K_2O}=a\left(mol\right);n_{KOH}=b\left(mol\right)\)
=> \(x=m_{K_2O}=94a\left(g\right)\)
=> \(m_{KOH}=56b\left(g\right)\Rightarrow y=m_{dd.KOH}=\dfrac{56b}{12\%}=\dfrac{1400b}{3}\left(g\right)\)
=> \(m_{dd.KOH.mới}=94a+\dfrac{1400b}{3}=120\left(g\right)\left(1\right)\)
PTHH: \(K_2O+H_2O\rightarrow2KOH\)
a------------------>2a
=> \(m_{KOH\left(sinh.ra\right)}=56.2a=112a\left(g\right)\)
\(m_{KOH\left(trong.dd.mới\right)}=112a+56b=51\%.120=61,2\left(g\right)\left(2\right)\)
Từ (1), (2) => \(\left\{{}\begin{matrix}a=0,4646\left(mol\right)\\b=0,1635\left(mol\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=94.0,4646=53,6724\left(g\right)\\y=\dfrac{1400}{3}.0,1635=76,3\left(g\right)\end{matrix}\right.\)
