Câu 6:
Ta có: \(n_{P_2O_5}=\dfrac{7,1}{142}=0,05\left(mol\right)\)
PT: \(P_2O_5+6KOH\rightarrow2K_3PO_4+3H_2O\)
____0,05____0,3_______0,1 (mol)
Ta có: m dd sau pư = 7,1 + 100 = 107,1 (g)
\(C\%_{K_3PO_4}=\dfrac{0,1.212}{107,1}.100\%\approx19,8\%\)
\(m_{KOH}=0,3.56=16,8\left(g\right)\)
\(\Rightarrow x=\dfrac{16,8}{100}.100\%=16,8\%\)
Bạn tham khảo nhé!