Answer:
Part A: 0.500 L
Part B: 1.220 M
Explanation:
Part A
The balanced reaction is:
2HCl(aq) + Na₂CO₃(aq) → 2NaCl(aq) + H₂O(l) + CO₂(g)
So, the
stoichiometry is 2 moles of HCl for 1 mol of Na₂CO₃. The number of moles (n) is the concentration multiplied by the volume, so:
n = 0.500x0.500 = 0.250 mol of Na₂CO₃.
So:
2 moles of HCl ----------------- 1 mol of Na₂CO₃
x ----------------- 0.250 mol of Na₂CO₃
By a simple direct three rule:
x = 0.500 mol
The volume will be the number of moles divided by the molar concentration:
V = 0.500/1.00 = 0.500 L
Part B
Now, the stoichiometry is 2 moles of HCl to form 1 mol of CO₂. We need to know how much moles of CO₂ are formed, knowing the molar masses:
MC = 12 g/mol
MO = 16 g/mol
MCO₂ = 12 +2x16 = 44 g/mol
The number of moles is the mass divided by the molar mass, so:
n = 11.1/44 = 0.252 mol
Then:
2 moles of HCl -------------------- 1 mol of CO₂
x -------------------- 0.252 mol of CO₂
By a simple direct three rule:
x = 0.504 mol of HCl
So, the concentration of HCl is the number of moles divided by the volume in liters ( 413 mL = 0.413 L):
0.504/0.413 = 1.220 M
