\(n_{NaOH}=0.2\cdot0.5=0.1\left(mol\right)\)
\(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)
\(0.1.......................0.1\)
\(m_{dd_{CH_3COOH}}=\dfrac{0.1\cdot60\cdot100}{5}=120\left(g\right)\)
=> B
$CH_3COOH + NaOH \to CH_3COONa + H_2O$
n CH3COOH = n NaOH = 0,2.0,5 = 0,1(mol)
=> m dd CH3COOH = 0,1.60/5% = 120(gam)
Đáp án B
Ta có: nNaOH = 0,2.0,5 = 0,1 (mol)
PT: \(NaOH+CH_3COOH\rightarrow CH_3COONa+H_2O\)
_____0,1_______0,1 (mol)
⇒ mCH3COOH = 0,1.60 = 6 (g)
\(\Rightarrow m_{ddCH_3COOH}=\dfrac{6}{5\%}=120\left(g\right)\)
⇒ Đáp án: B
Bạn tham khảo nhé!
