The approximate concentration of the NaOH solution you will prepare in Part A of this experiment is 0.12M. Calculate the number

Question

3 years ago

14

of moles and the number of grams of oxalic acid (H2C2O4 •2H20) that will be required to neutralize 25 mL of this NaOH solution.

2 answers:

zhuklara [117] · Answer 1 · 2022-08-23T12:11:28+03:00

Answer:

0.0015moles

0.135g

Explanation:

Given parameters:

Volume of NaOH = 25mL = 0.025L

Molarity of NaOH = 0.12M

Unkown:

Number of moles of oxalic acid = ?

mass of oxalic acid = ?

Solution:

The reaction equation is shown below:

H₂C₂O₄ + 2NaOH → Na₂C₂O₄ + 2H₂O

Solving:

We should work from the known to the unknown. The known here is NaOH.

##

Number of moles of NaOH = molarity of NaOH x volume = 0.12 x 0.025 = 0.003mole

##

From the equation:

2 moles of NaOH reacted with 1 mole of the acid

0.03 mole of NaOH will react with $\frac{0.003}{2}$ = 0.0015moles of oxalic acid

##

Mass of oxalic acid = number of moles x molar mass

Molar mass of oxalic acid = (2 x 1) + (2 x 12) + (4 x 16) = 90g/mol

Mass of oxalic acid = 0.0015 x 90 = 0.135g of oxalic acid.

aev [14] · Answer 2 · 2022-08-23T14:31:00+03:00

Answer:

number of grams of oxalic acid = 0.135 g

Explanation:

Thinking process:

The reaction equation of hydrated oxalic acid is shown below:

H₂C₂O₄ + 2NaOH → Na₂C₂O₄ + 2H₂O

number of moles of NaOH = concatenation × volume

concentration of NaOH = 0.12M

volume of NaOH = 25 ml

= 0.025 l

The number of moles = $0.025 * 0.12\\= 0.003 moles$

The mole ratio of oxalic acid is half the number of moles of NaOH = $0.5 * 0.003\\= 0.0015 mol$

Molar mass of oxalic acid = 90.03

number of moles = $\frac{mass}{molar mass}$

mass = $number of moles * molar mass\\ = 0.0015 * 90.03\\ = 0.135 g$