How many grams of benzoic acid would be required to react with the naoh sample mention in question 2

The question is incomplete, the complete question is;

2. How many moles of NaOH are there in a 20.0mL sample of an 0.101M solution?

3. How many grams of benzoic acid would be required to react with the NaOH

sample mention in Question 2?

0.247 g

Explanation:

Recall that;

n =CV

n = number of moles

C = concentration

V = volume (in litres)

So;

n = 0.101 * 20/1000

n = 2.02 * 10^-3 moles

The equation of the reaction is;

C6H5COOH(s)+NaOH(aq) > C6H5COO-Na+ (aq)+H2O(l)

Given that the reaction is in 1:1 molar ratio

number of moles of benzoic acid required= 2.02 * 10^-3 moles * 122.12 g/mol = 0.247 g

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"

the answer is minerals.