\(n_{HCOOH} = a(mol) ; n_{CH_3COOH} =b(mol)\\ \Rightarrow 46a + 60b = 15,2(1)\\ HCOOH + NaHCO_3 \to HCOONa + CO_2 + H_2O\\ CH_3COOH + NaHCO_3 \to CH_3COONa + CO_2 + H_2O\\ n_{CO_2} = a + b = \dfrac{3,36}{22,4} = 0,15(2)\\ (1)(2) \Rightarrow a = \dfrac{-31}{70}<0\)
(Sai đề)
Sửa đề : Na2CO3
\(n_{HCOOH}=a\left(mol\right),n_{CH_3COOH}=b\left(mol\right)\)
\(\Rightarrow m_{hh}=46a+60b=15.2\left(g\right)\left(1\right)\)
\(2HCOOH+Na_2CO_3\rightarrow2HCOONa+CO_2+H_2O\)
\(a......................................0.5a\)
\(2CH_3COOH+Na_2CO_3\rightarrow2CH_3COONa+CO_2+H_2O\)
\(b...........................................0.5b\)
\(n_{CO_2}=0.5a+0.5b=\dfrac{3.36}{22.4}=0.15\left(mol\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.2,b=0.1\)
\(m_{HCOOH}=0.2\cdot46=9.2\left(g\right)\)
\(m_{CH_3COOH}=6\left(g\right)\)
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