What will happen if you add more solute to a saturated solution?
1 answer:
You might be interested in
Answer:
<em> = 0.2 mL.</em>
Explanation:
Given a 0.5 M solution of NaOH as stock solution, 10.0mL of 0.010M can be prepared via dilution with distilled water, by using the formula:
where C1 and V1 are initial concentration and volume respectively; same as C2 & V2 for fina.
Let C1 = 0.5M, V2 = ?
C2 = 0.010M; V2 = 10mL
⇒Volume of stock solution to be diluted, V2
= × 0.010
<em> = 0.2 mL.</em>
Glasswares used would be pipette (for smaller volume experiment) and measuring cylinder. 0.2mL would be measured and then made upto the 10mL mark of the measuring cylinder.
I hope this was a detailed explanation given the missing details of "Trial 1" in the question.
Answer:- As per the question is asked, 35.0 moles of acetylene gives 70 moles of carbon dioxide but if we solve the problem using the limiting reactant which is oxygen then 67.2 moles of carbon dioxide will form.
Solution:- The balanced equation for the combustion of acetylene is:
From the balanced equation, two moles of acetylene gives four moles of carbon dioxide. Using dimensional analysis we could show the calculations for the formation of carbon dioxide by the combustion of 35.0 moles of acetylene.
=
The next part is, how we choose 35.0 moles of acetylene and not 84.0 moles of oxygen.
From balanced equation, there is 2:5 mol ratio between acetylene and oxygen. Let's calculate the moles of oxygen required to react completely with 35.0 moles of acetylene.
=
Calculations shows that 87.5 moles of oxygen are required to react completely with 35.0 moles of acetylene. Since only 84.0 moles of oxygen are available, the limiting reactant is oxygen, so 35.0 moles of acetylene will not react completely as it is excess reactant.
So, the theoretical yield should be calculated using 84.0 moles of oxygen as:
=