Final answer:
The mass of sodium sulfate (Na2SO4) formed when 0.5 moles of sodium hydroxide react is 35.5 grams. This is calculated using the balanced chemical equation and the molar mass of Na2SO4.
Step-by-step explanation:
The mass of sodium tetraoxosulphate (VI) or sodium sulfate (Na2SO4) formed when 0.5 moles of sodium hydroxide (NaOH) react can be determined by first writing the balanced chemical equation for the reaction between sodium hydroxide and sulfuric acid (assuming 'tetraoxic acid' refers to sulfuric acid, H2SO4):
H2SO4 (aq) + 2NaOH (aq) → Na2SO4 (aq) + 2H2O(l)
From the equation, it's clear that 2 moles of NaOH react with 1 mole of H2SO4 to form 1 mole of Na2SO4. Therefore, 0.5 moles of NaOH would react to form 0.25 moles of Na2SO4.
To calculate the mass of Na2SO4 formed:
Molar mass of Na2SO4 = 2(Na) + 1(S) + 4(O) = 2(23 g/mol) + 32 g/mol + 4(16 g/mol)
= 142 g/mol
Mass of Na2SO4 = moles × molar mass
= 0.25 moles × 142 g/mol
= 35.5 grams.