First, let's compute the number of moles in the system assuming ideal gas behavior.
PV = nRT
(663 mmHg)(1atm/760 mmHg)(60 L) = n(0.0821 L-atm/mol-K)(20+273 K)
Solving for n,
n = 2.176 moles
At standard conditions, the standard molar volume is 22.4 L/mol. Thus,
Standard volume = 22.4 L/mol * 2.176 mol =<em> 48.74 L</em>
Answer:
5.158 mol/L
Explanation:
To find the molarity, you need to use the formula:
Molarity (M) = moles / volume (L)
You have been grams sodium carbonate. You need to (1) convert grams Na₂CO₃ to moles (via molar mass), then (2) convert moles Na₂CO₃ to moles HCl (via mole-to-mole ratio from equation), then (3) convert mL to L (by dividing by 1,000), and then (4) use the molarity equation.
<u>Steps 1 - 2:</u>
2 HCl + 1 Na₂CO₃ ----> 2 NaCl + H₂O + CO₂
6.5287 g Na₂CO₃ 1 mole 2 moles HCl
-------------------------- x ------------- x ------------------------- = 0.12318 mole HCl
106 g 1 mole Na₂CO₃
<u>Step 3:</u>
23.88 mL / 1,000 = 0.02388 L
<u>Step 4:</u>
Molarity = moles / volume
Molarity = 0.12318 mole / 0.02388 L
Molarity = 5.158 mole/L
**mole/L is equal to M**
Mass of KCl= 1.08 g
<h3>Further explanation</h3>
Given
1 g of K₂CO₃
Required
Mass of KCl
Solution
Reaction
K₂CO₃ +2HCl ⇒ 2KCl +H₂O + CO₂
mol of K₂CO₃(MW=138 g/mol) :
= 1 g : 138 g/mol
= 0.00725
From the equation, mol ratio K₂CO₃ : KCl = 1 : 2, so mol KCl :
= 2/1 x mol K₂CO₃
= 2/1 x 0.00725
= 0.0145
Mass of KCl(MW=74.5 g/mol) :
= mol x MW
= 0.0145 x 74.5
= 1.08 g
