Sodium hydroxide reacted with sulphuric (6) acid to form 8g of a salt . Calculate the mass of sodium hydroxide that reacted
To solve this problem, we need to use stoichiometry and the balanced chemical equation between sodium hydroxide (NaOH) and sulfuric acid (H2SO4).
The balanced equation is:
2NaOH + H2SO4 -> Na2SO4 + 2H2O
From the equation, we can see that for every 2 moles of NaOH, we need 1 mole of H2SO4 to react. This means that the mole ratio between NaOH and H2SO4 is 2:1.
To find the number of moles of NaOH that reacted, we need to calculate the number of moles of the salt (Na2SO4) produced. We can use the molar mass of Na2SO4 to convert the mass of the salt to moles.
Molar mass of Na2SO4 = (2 x atomic mass of Na) + atomic mass of S + (4 x atomic mass of O)
= (2 x 22.99 g/mol) + 32.07 g/mol + (4 x 16.00 g/mol)
= 46.00 g/mol + 32.07 g/mol + 64.00 g/mol
= 142.07 g/mol
Number of moles of Na2SO4 = mass of salt / molar mass of Na2SO4
= 8g / 142.07 g/mol
= 0.0562 mol
Since the mole ratio between NaOH and Na2SO4 is 2:1, the number of moles of NaOH that reacted is half the number of moles of Na2SO4.
Number of moles of NaOH = 0.0562 mol / 2
= 0.0281 mol
To calculate the mass of NaOH that reacted, we can use the molar mass of NaOH.
Molar mass of NaOH = atomic mass of Na + atomic mass of O + atomic mass of H
= 22.99 g/mol + 16.00 g/mol + 1.01 g/mol
= 40.00 g/mol
Mass of NaOH that reacted = number of moles of NaOH x molar mass of NaOH
= 0.0281 mol x 40.00 g/mol
= 1.124 g
Therefore, the mass of sodium hydroxide that reacted is 1.124 grams.