a. 43.1 g
b. 38.2%
<h3>Further explanation</h3>
Given
32.5 grams of NaOH
Required
The theoretical yield of Na₂CO₃
The percent yield
Solution
Reaction
2NaOH(s) + CO₂(g) → Na₂CO₃(s) + H₂O(l)
mol NaOH :
= mass : MW
= 32.5 : 40 g/mol
= 0.8125
mol Na₂CO₃ from the equation :
= 1/2 x mol NaOH
= 1/2 x 0.8125
= 0.40625
a.
Mass Na₂CO₃ :
= mol x MW Na₂CO₃
= 0.40625 x 106 g/mol
= 43.0625≈43.1 g
b. % yield = (actual/theoretical) x 100%
%yield = 16.45/43.1 x 1005
%yield = 38.17%≈38.2%
Answer is 10 units of water molecules.
Answer:
4.33 L
Explanation:
Assuming ideal behaviour and that all 0.300 moles of gas reacted, we can solve this problem using Avogadro's law, which states that at constant temperature and pressure:
Where in this case:
We <u>input the given data</u>:
- 2.16 L * 0.601 mol = V₂ * 0.300 mol
And <u>solve for V₂</u>:
The answer is Latitude (B)
