In the same reaction hcl (aq) + naoh (aq) → nacl (aq) + h2o (l), if 125.0 milliliters of a 1.5 m hcl react in with 87.5 milliliters of 1.75 m naoh, how many grams of h2o will be produced?

1) Chemical reaction

HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l),

2) Molar ratios

1 mol HCl : 1 mol NaOH : 1 mol NaCl : 1 mol H2O

3) Find the number of moles of each reactant

125.0 milliliters of a 1.5 M HCl

M = n / V => n = M * V = 1.5 M * 0..1250 l = 0.1875 mol HCl

87.5 milliliters of 1.75 M NaOH,

n = M*V = 1.75 M * 0.0875 l = 0.1531 mol NaOH

3) Limiting reagent

It is NaOH because its 0.1531 moles will be consumed before the 0.1875 mol of HCl are depleted.

4) Use molar ratios to determine the number of moles of H2O produced:

1:1 => 0.1531 mole of H2O

5) how many grams of H2O will be produced?

use the molar mass to convert 0.1531 mol of H2O to grams:

0.1531 mol * 18.0 g / mol = 2.76 g

2.76 g

☆ionisation energy increases across a period from na -> ar☆ionisation energy decreases down a group from li -> fr☆ionic radius decreases across a period from na -> cl☆ionic radius increases down a groupfrom f -> ii hope it clarifies your doubts