Question

How many grams of oxalic acid dihydrate are needed to react with 20 ml of 0.45 M NaOH?

A) 0.9 g
B) 1.8 g
C) 3.6 g
D) 5.4 g

Answer 1

To react with 20 mL of 0.45 M NaOH, you would need 0.567 g of oxalic acid dihydrate.

Step-by-step explanation:

In order to determine the amount of oxalic acid dihydrate needed to react with 20 mL of 0.45 M NaOH, we need to use the balanced chemical equation for the reaction. The equation shows that the stoichiometric ratio between NaOH and oxalic acid dihydrate is 2:1. This means that for every 2 moles of NaOH, we need 1 mole of oxalic acid dihydrate.

To calculate the grams of oxalic acid dihydrate needed, we can use the molar mass of oxalic acid dihydrate and the molarity of NaOH.

Given that the molar mass of oxalic acid dihydrate is 126.07 g/mol, we can use the following calculation:

1. Calculate the moles of NaOH: 0.45 M x 0.02 L = 0.009 moles
2. Calculate the moles of oxalic acid dihydrate: 0.009 moles x (1 mole/2 moles) = 0.0045 moles
3. Calculate the grams of oxalic acid dihydrate: 0.0045 moles x 126.07 g/mol = 0.567 g