Answer:
Molarity of HCl=0.026092M
Explanation:
The equation for the reaction is;
HCl + NaOH ⇒ NaCl + H2O
Using the formular, ..........equ1
where is the concentration of Acid,
is volume of acid
is concentration of the base
is volume of the base
nA is the number of moles of Acid
nB is number of moles of base
nA = 1, nB=1 , ,
,
we will input these values into equation1 to solve for
Answer:
(B) 0.160 M
Explanation:
Considering:
Or,
Given :
For :
Molarity = 0.200 M
Volume = 20.0 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 20.0×10⁻³ L
Thus, moles of :
Moles of = 0.004 moles
For :
Molarity = 0.400 M
Volume = 30.0 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume =30.0×10⁻³ L
Thus, moles of :
Moles of = 0.012 moles
According to the given reaction:
1 mole of potassium carbonate react with 1 mole of barium nitrate
0.004 moles potassium carbonate react with 0.004 mole of barium nitrate
Moles of barium nitrate = 0.004 moles
Available moles of barium nitrate = 0.012 moles
Limiting reagent is the one which is present in small amount. Thus, potassium carbonate is limiting reagent.
The formation of the product is governed by the limiting reagent. So,
1 mole of potassium carbonate gives 1 mole of barium carbonate
Also,
0.004 mole of potassium carbonate gives 0.004 mole of barium carbonate
Mole of barium carbonate = 0.004 moles
Also, consumed barium nitrate = 0.004 moles (barium ions precipitate with carbonate ions)
Left over moles = 0.012 - 0.004 moles = 0.008 moles
Total volume = 20.0 + 30.0 mL = 50.0 mL = 0.05 L
So, Concentration = 0.008/0.05 M = 0.160 M
<u>(B) is correct.</u>