Answer :
The maximum mass of sodium sulfate that could be produced by the chemical reaction is 125 g.
To calculate the maximum mass of sodium sulfate produced, we need to determine the limiting reactant in the reaction. This can be done by comparing the moles of sulfuric acid and sodium hydroxide.
First, let's convert the given masses of sulfuric acid and sodium hydroxide to moles. The molar mass of H2SO4 is 98.09 g/mol, and the molar mass of NaOH is 39.997 g/mol.
For sulfuric acid:
97.9 g H2SO4 × (1 mol H2SO4 / 98.09 g H2SO4) = 0.998 mol H2SO4
For sodium hydroxide:
44.3 g NaOH × (1 mol NaOH / 39.997 g NaOH) = 1.108 mol NaOH
Next, we need to determine the mole ratio between sulfuric acid and sodium sulfate from the balanced chemical equation:
H2SO4 + 2NaOH → Na2SO4 + 2H2O
From the equation, we can see that 1 mole of sulfuric acid reacts with 1 mole of sodium sulfate. Therefore, the maximum moles of sodium sulfate that can be produced is equal to the moles of sulfuric acid, which is 0.998 mol.
Finally, we convert the moles of sodium sulfate to grams using the molar mass of Na2SO4 (142.04 g/mol):
0.998 mol Na2SO4 × (142.04 g Na2SO4 / 1 mol Na2SO4) = 141.9 g Na2SO4
Rounding to three significant digits gives us a maximum mass of 125 g for sodium sulfate.
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