Question

cho 100g Ba(OH)₂. Tính số ml HCl 0.5M cần dùng để trung hòa hết 100g Ba(OH)₂ đó

ai giải hộ vs

Answer 1

\(n_{Ba\left(OH\right)_2}=\dfrac{100}{171}\left(mol\right)\)

\(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)

\(\dfrac{100}{171}........\dfrac{200}{171}\)

\(V_{dd_{HCl}}=\dfrac{\dfrac{200}{171}}{0.5}=2.34\left(l\right)=2340\left(ml\right)\)

Answer 2

\(n_{Ba(OH)_2}=\dfrac{100}{171}\approx 0,58(mol)\\ Ba(OH)_2+2HCl\to BaCl_2+2H_2O\\ \Rightarrow n_{HCl}=2n_{Ba(OH)_2}=1,16(mol)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{1,16}{0,5}=2,32(l)=2320(ml)\)