calculate how many moles of the product form when 2.57 mol of Na completely reacts.

Assume that there is more than enough of the other reactant.
2Na(s)+O2(g)→Na2O2(s)

The coefficients tell you that.

2.57 molx Na x (1 mol Na2O2/2 mols Na) = ?
Note how mols Na in the numerator cancels with the mols Na in the denominator. That leaves mols Na2O2 in the numerator as the only unit left; you have converted mols Na to mols Na2O2.

To calculate the number of moles of the product formed when 2.57 mol of Na completely reacts, we need to use the balanced chemical equation and the mole ratio between the reactant (Na) and the product (Na2O2).

The balanced chemical equation is:
2Na(s) + O2(g) → Na2O2(s)

According to the equation, for every 2 moles of Na that reacts, 1 mole of Na2O2 is produced.

Therefore, we can use the mole ratio to calculate the number of moles of Na2O2 produced:

Moles of Na2O2 = (Moles of Na) * (1 mole of Na2O2 / 2 moles of Na)

Moles of Na2O2 = 2.57 mol * (1 mole of Na2O2 / 2 moles of Na)
Moles of Na2O2 = 1.285 mol

So, when 2.57 mol of Na completely reacts, 1.285 mol of Na2O2 is formed.

To calculate the number of moles of the product formed, we need to use the balanced chemical equation and the stoichiometry of the reaction.

The balanced chemical equation is:

2 Na(s) + O2(g) → Na2O2(s)

The coefficient in front of Na2O2 is 1, which means that for every 2 moles of Na reacting, we get 1 mole of Na2O2.

We are given that 2.57 mol of Na completely reacts. From the balanced equation, we can see that this corresponds to half the molar amount of Na2O2 formed.

To calculate the moles of Na2O2:

2.57 mol Na × (1 mol Na2O2 / 2 mol Na) = 1.285 mol Na2O2

Therefore, when 2.57 mol of Na completely reacts, it forms 1.285 mol of Na2O2.