Acetone to Sodium Hypochlorite Ratio for Chloroform Production: A Comprehensive Guide
Understanding the precise ratio of acetone to sodium hypochlorite is crucial for the production of chloroform from bleach. This process involves a series of calculations and chemical reactions. This guide breaks down the detailed steps, including the ratio of acetone to sodium hypochlorite, to ensure accurate chloroform synthesis.
What is Sodium Hypochlorite (Bleach)?
Bleach, typically aqueous sodium hypochlorite (NaClO), is an essential component in this process. Sodium hypochlorite has a molar mass of 74.45 g/mol, derived from its constituent elements: 23 g/mol for sodium (Na), 35.45 g/mol for chlorine (Cl), and 16 g/mol for oxygen (O).
Calculating the Active Chlorine Content
The active chlorine content in bleach can be calculated by considering the absence of sodium chloride (NaCl) waste. For simplicity, we assume bleach has a concentration of 7% active chlorine (Cl) by mass. This means that in a 100 g solution, 7 g is chlorine and the remaining 93 g is water and sodium hypochlorite (NaClO).
The calculation for the moles of sodium hypochlorite (NaClO) in a 7% solution can be done as follows:
Active chlorine (Cl) mass 7 g Molar mass of NaClO 74.45 g/mol Molar mass of water (H2O) 18 g/mol The equation for the ratio of moles is:[frac{n_{text{mol}} times 74.45,text{g/mol}}{k times 18,text{g/mol} n_{text{mol}} times 74.45,text{g/mol}} 0.07]
Solving this equation, we find that 1 mole of NaClO requires approximately 24 moles of water to maintain the 7% active chlorine content.
Moles and Mass of Sodium Hypochlorite and Water
Given 1 liter of water (1,000 g) with a 7% active chlorine content, the mass of sodium hypochlorite can be calculated:
Moles of water (frac{1,000,text{g}}{18,text{g/mol}} 55.56,text{mol}) Moles of sodium hypochlorite (frac{55.56}{24},text{mol} 2.315,text{mol}) Mass of sodium hypochlorite (2.315,text{mol} times 74.45,text{g/mol} 172.35,text{g}) Total solution mass water sodium hypochlorite 1,000 g 172.35 g 1,172.35 gReaction Between Acetone and Sodium Hypochlorite
The balanced chemical reaction for the production of chloroform from acetone and sodium hypochlorite is as follows:
(text{C}_3text{H}_6text{O} 3text{NaClO} rightarrow text{CHCl}_3 2text{NaOH} text{NaCH}_3text{COO})
This reaction shows that acetone (C3H6O) and sodium hypochlorite (NaClO) react in a 1:3 molar ratio. Therefore, for every 1 mole of acetone, 3 moles of sodium hypochlorite are needed.
Calculating the Mass of Acetone
The molar mass of acetone (C3H6O) is:
(3 times 12,text{g/mol} 6 times 1,text{g/mol} 16,text{g/mol} 58,text{g/mol})
Given a 7% active chlorine solution, the moles of acetone required would be:
(0.77,text{mol})
The corresponding mass of acetone would then be:
(0.77,text{mol} times 58,text{g/mol} 44.66,text{g})
Complete Mass Ratio for Chloroform Production
The complete mass ratio of acetone to the 7% active chlorine bleach solution can be calculated as follows:
(frac{44.66,text{g}}{1,172.35,text{g}} 0.0381)
The inverse of this ratio (bleach to acetone) is approximately 26.25. This means that for every 1 g of acetone, 26.25 g of the 7% active chlorine bleach solution are needed.
Volume Considerations
Considering the densities of acetone (784 g/L) and the bleach solution (approximately 1,172.35 g/L), the volume ratio of acetone to bleach can be calculated. Given the density difference:
(frac{44.66,text{g}}{784,text{g/L}} 0.057)
Therefore, the volume ratio of acetone to bleach is approximately 1:17.55.
The detailed calculations and considerations provide a clear understanding of the acetone to sodium hypochlorite ratio for chloroform production, ensuring the effectiveness and efficiency of the chemical process.