Question

Який мінімальний об’єм амоніаку потрібно пропустити крізь розчин масою 300г з масовою часткою хлороцтової кислоти 20% для повного перетворення її на амінооцтову кислоту?

Answer 1

Ответ:28,5

Объяснение:

Answer 2

I will answer in detail. The minimum volume of ammonia required to pass through a solution with a mass of 300 g and a mass fraction of chloroacetic acid 20% for complete conversion to aminoacetic acid is 14.2 liters. Here is how I calculated it:

- First, I found the pure mass of the acid: m (acid) = m (solution) * w (acid) = 300 * 20% / 100% = 60 g - Then, I found the molar mass of the acid: M (ClCH2COOH) = 35.5 + 12 + 1 * 2 + 12 + 16 + 16 + 1 = 94.5 g/mol - Next, I wrote the chemical equation for the reaction: ClCH2COOH + NH3 -> NH2CH2COOH + HCl - I balanced the equation by adding coefficients: ClCH2COOH + NH3 -> NH2CH2COOH + HCl - I calculated the number of moles of the acid: n (acid) = m (acid) / M (acid) = 60 / 94.5 = 0.635 mol - I used the mole ratio to find the number of moles of ammonia: n (NH3) = n (acid) * 1 / 1 = 0.635 mol - I used the molar volume of ammonia at normal conditions to find the volume of ammonia: V (NH3) = n (NH3) * 22.4 L/mol = 0.635 * 22.4 = 14.2 L

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