Question

1.Cho 40 g CaCO₃ tác dụng với V (l) dd axit axetic 1M, phản ứng kết thúc tạo thành 4,48 l khí CO2 (đktc). tính V?

Answer 1

nCO2 = 4.48/22.4 = 0.2 (mol) 

2CH3COOH  + CaCO3 => (CH3COO)2Ca + CO2 + H2O 

0.4..........................0.2......................................0.2

V_CH3COOH = 0.4/1 = 0.4 (l) 

Answer 2

PTHH: \(CaCO_3+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Ca+H_2O+CO_2\uparrow\)

Ta có: \(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\) \(\Rightarrow n_{CH_3COOH}=0,4\left(mol\right)\)

\(\Rightarrow V_{CH_3COOH}=\dfrac{0,4}{1}=0,4\left(l\right)=400\left(ml\right)\)

Answer 3

\(CaCO_3 + 2CH_3COOH \to (CH_3COO)_2Ca + CO_2 + H_2O\\ n_{CaCO_3} = \dfrac{40}{100} = 0,4 > n_{CO_2} = \dfrac{4,48}{22,4} = 0,2 \to CaCO_3\ dư\\ n_{CH_3COOH} = 2n_{CO_2} = 0,2.2 = 0,4(mol)\\ \Rightarrow V_{CH_3COOH} = \dfrac{0,4}{1} = 0,4(lít)\)