Answer:
The mass of sodium sulfate formed by the comolete reaction of 120.0 grams of sodium hydroxide is 142.04 grams.
Step-by-step explanation:
The balanced chemical equation for the reaction between sodium hydroxide and sulfuric acid is:
2NaOH + H2SO4 -> Na2SO4 + 2H2O
From the equation, we can see that 2 moles of sodium hydroxide react with 1 mole of sulfuric acid to form 1 mole of sodium sulfate. We can use this information, along with the molar masses of the compounds, to calculate the mass of sodium sulfate formed.
First, we need to convert the given mqss of sodium hydroxide to moles. The molar mass of sodium hydroxide is 40.00 g/mol, so:
Moles of NaOH = Mass of NaOH / Molar mass of NaOH
Moles of NaOH = 120.0 g / 40.00 g/mol
Moles of NaOH = 3.00 mol
Next, we can use the mole ratio from the balanced equation to calculate the moles of sodium sulfate formed:
Moles of Na2SO4 = Moles of NaOH / 2
Moles of Na2SO4 = 3.00 mol / 2
Moles of Na2SO4 = 1.50 mol
Finally, we can convert the moles of sodium sulfate to grams using its molqr mass of 142.04 g/mol:
Mass of Na2SO4 = Moles of Na2SO4 x Molar mass of Na2SO4
Mass of Na2SO4 = 1.50 mol x 142.04 g/mol
Mass of Na2SO4 = 213.06 g
Therefore, the mass of sodium sulfate formed by the complete reaction of 120.0 grams of sodium hydroxide is 213.06 grams.