Question

Đem trung hòa 600ml dd HCl 1M bằng dd NaOH 30%.

a) Tính m dd NaOH đã dùng.

b) Nếu thay dd NaOH 30% bằng dd Ca(OH)2 7.351% ( D= 1.045g/ml ) thì V dd Ca(OH)2 cần dùng là bnh?

Answer 1

n_HCl=C_M.V=1.0,6=0,6(mol)

Pt: HCl+ NaOH-> NaCl + H₂O

cứ:1.............1.............1 (mol)

vậy:0,6----->0,6------>0,6(mol)

=> m_NaOH=0,6.40=24(g)

\(\Rightarrow m_{ddNaOH}=\dfrac{m_{NaOH}.100\%}{C\%}=\dfrac{24.100}{30}=80\left(g\right)\)

b) Nếu thay NaOH bằng Ca(OH)₂ thì ta có PT:

PT: 2HCl + Ca(OH)₂ -> CaCl₂ +2H₂O

Cứ:2................1.................1 (mol)

Vậy: 0,6------->0,3----------->0,3(mol)

=> m_Ca(OH)2=n.M=0,3.74=22,2(g)

\(\Rightarrow m_{ddCa\left(OH\right)_2}=\dfrac{m_{Ca\left(OH\right)_2}.100\%}{C\%}=\dfrac{22,2.100}{7,351}\approx302\left(g\right)\)

\(\Rightarrow V_{ddCa\left(OH\right)_2}=\dfrac{m_{ddCa\left(OH\right)_2}}{D}=\dfrac{302}{1,045}\approx289\left(ml\right)=0,289\left(lít\right)\)