Hi! where are the options?
Answer:
The compound you will use is the Dibasic phosphate
Explanation:
Simple phosphate buffer is used ubiquitously in biological experiments, as it can be adapted to a variety of pH levels, including isotonic. This wide range is due to phosphoric acid having 3 dissociation constants, (known in chemistry as a triprotic acid) allowing for formulation of buffers near each of the pH levels of 2.15, 6.86, or 12.32. Phosphate buffer is highly water soluble and has a high buffering capacity,
In this case the most efficient way is to disolve the dibasic compound which in the reaction with the water will form the monobasic phosphate.
To make the buffer you have to prepare the amount of distillate water needed, disolve the dibasic phospate, and then adjust with HCl or NaOH depending on the pH needed.
Answer:
25.53mL of 0.200 M FeCl₃ are needed to produce 0.345g of Fe₂S₃
Explanation:
Based on the reaction of the problem, 1 mole of Fe₂S₃ is produced from 2 moles of FeCl₃.
0.345g of Fe₂S₃ are (Molar mass: 207.9g/mol):
0.345g of Fe₂S₃ ₓ (1 mol / 207.9g) = <em>1.6595x10⁻³ moles Fe₂S₃</em>
Moles of Fe needed to produce these moles of Fe₂S₃ are:
1.6595x10⁻³ moles Fe₂S₃ ₓ ( 2 moles FeCl₃ / 1 mole Fe₂S₃) =
<em>3.3189x10⁻³ moles of FeCl₃</em>
As the percent yield of the reaction is 65.0%, the moles of FeCl₃ you need to add are:
3.3189x10⁻³ moles of FeCl₃ ₓ (100.0% / 65.0%) = <em>5.106x10⁻³ moles of FeCl₃</em>
A solution 0.200M contains 0.200 moles per L. Volume to obtain 5.106x10⁻³ moles is:
<em>5.106x10⁻³ moles of FeCl₃ ₓ ( 1L / 0.200mol) = 0.02553L = </em>
<h3>25.53mL of 0.200 M FeCl₃ are needed to produce 0.345g of Fe₂S₃</h3>
The reactants are found on the left side, the products are found on the right side.
