Theo bài ra ta có:
+)mBa(OH)2 = \(\dfrac{C\%.mdd}{100\%}\) = \(\dfrac{10\%.855}{100\%}\)= 85,5 g
=> nBa(OH)2 = \(\dfrac{\text{mBa(OH)2 }}{\text{MBa(OH)2 }}\)= \(\dfrac{85,5}{171}\)= 0,5 mol
PTHH. Ba(OH)2 + H2SO4 -> 2H2O + BaSO4↓ (1)
0,5-----------0,5
Theo bài ra ta lại có:
+) mddNaOH = 125. 1,28 = 160g
=> mNaOH = \(\dfrac{C\%.mdd}{100\%}\) = \(\dfrac{25\%.160}{100\%}\) = 40g
=> nNaOH = \(\dfrac{\text{mNaOH }}{\text{MNaOH }}\)= \(\dfrac{40}{40}\) = 1 mol
PTHH. H2SO4 + 2NaOH --->Na2SO4 + 2H2O (2)
0,5-------1
Từ trên => nH2SO4 = nH2SO4(pt1) + nH2SO4(pt2)
= 0,5 + 0,5 = 1 mol
Đổi 200ml = 200g
=> mH2SO4 = nH2SO4 . MH2SO4 = 1. 98 = 98 g
=> C%ddH2SO4 = \(\dfrac{m_{ct}.100\%}{m_{dd}}\) = \(\dfrac{98.100\%}{200}\) = 49%
Vậy....
