If 27.37ml of 0.2115M NaOH is able to neutralize 37.45 of HCl, what is the concentration of the acid?

You just gotta match the moles:

moles of NaOH = 0.02737 L * 0.2115 M = 0.005789 mol of NaOH. The amount of moles of HCl it can neutralize should be the same since they are in a 1:1 ratio. Therefore, you can do moles of NaOH divided by 0.003745L to get the answer. It turns out to be 0.1546M.

Oxygen

0,108 moles

0g Li; 5,25g of O₂; 3,23g of Li₂O

Explanation:

The reaction is:

4Li + O₂ → 2Li₂O

a) Moles of lithium in 1,50g are:

1,50g × (1mol / 6,941g) = 0,216 moles of Li

Moles of O₂ are:

6,98g × (1mol / 32g) = 0,218 moles of O₂

The complete reaction of oxygen requires:

0,218 mol ₓ (4mol Li / 1 mol O₂) = 0,872 moles of Li.

As there are more moles of oxygen than required moles of lithium the reactant in excess is oxygen

b) The limiting reactant is lithium, that means moles of product formed are moles of lithium ÷2, that is, 0,108 moles

a) After the reaction there are 0 moles of Li, 0,218 - 0,216× (1 mol O₂/4 mol Li) = 0,164 moles of O₂ and 0,432 moles of Li₂O. In grams:

0g Li

0,164 mol ₓ (32g / mol) = 5,25g of O₂

0,108mol Li₂O ₓ (29,88g / mol) = 3,23g of Li₂O

To know your answer is right you can sum the initial total mass and must be the same of final total mass

I hope it helps!