Question

hòa tan 50g dd Ba(OH)2 17,1% vào ddd K2CO3 13,8%

a) Viết PTHH

b) Tính C% dd sau PƯ

Answer 1

\(a.Ba\left(OH\right)_2+K_2CO_3\rightarrow BaCO_3+2KOH\\ n_{Ba\left(OH\right)_2}=\dfrac{50.17,1}{171}=0,05\left(mol\right)\\ n_{K_2CO_3}=n_{BaCO_3}=n_{Ba\left(OH\right)_2}=0,05\left(mol\right)\\ n_{KOH}=2n_{Ba\left(OH\right)_2}=0,1\left(mol\right)\\ m_{ddK_2CO_3}=\dfrac{0,05.138}{13,8\%}=50\left(g\right)\\ m_{ddsaupu}=50+50-0,05.197=90,15\left(g\right)\\ C\%_{KOH}=\dfrac{0,1.56}{90,15}.100=6,21\%\)