Aqueous sulfuric acid
H2SO4
will react with solid sodium hydroxide
NaOH
to produce aqueous sodium sulfate
Na2SO4
and liquid water
H2O
. Suppose 31. g of sulfuric acid is mixed with 11.5 g of sodium hydroxide. Calculate the maximum mass of sodium sulfate that could be produced by the chemical reaction. Round your answer to
3
significant digits.
.43 g
.431 g
To calculate the maximum mass of sodium sulfate that could be produced, we need to determine the limiting reagent. The limiting reagent is the reactant that is entirely consumed during the reaction, limiting the amount of product that can be formed.
To find the limiting reagent, we need to compare the given masses of sulfuric acid (H2SO4) and sodium hydroxide (NaOH) with their respective molar masses.
1. Calculate the molar mass of sulfuric acid (H2SO4):
- The atomic masses of hydrogen (H), sulfur (S), and oxygen (O) are approximately 1 g/mol, 32 g/mol, and 16 g/mol, respectively.
- Molar mass of H2SO4 = 2(1 g/mol) + 32 g/mol + 4(16 g/mol) = 98 g/mol
2. Calculate the molar mass of sodium hydroxide (NaOH):
- The atomic masses of sodium (Na), oxygen (O), and hydrogen (H) are approximately 23 g/mol, 16 g/mol, and 1 g/mol, respectively.
- Molar mass of NaOH = 23 g/mol + 16 g/mol + 1 g/mol = 40 g/mol
3. Determine the moles of sulfuric acid (H2SO4):
- Moles = Mass / Molar mass = 31 g / 98 g/mol ≈ 0.316 mol
4. Determine the moles of sodium hydroxide (NaOH):
- Moles = Mass / Molar mass = 11.5 g / 40 g/mol = 0.2875 mol
5. Calculate the stoichiometric ratio between sulfuric acid (H2SO4) and sodium sulfate (Na2SO4) using the balanced chemical equation:
- Balanced equation: 2 H2SO4 + 2 NaOH → Na2SO4 + 2 H2O
- From the equation, we can see that 2 moles of sulfuric acid react with 1 mole of sodium sulfate.
6. Determine the limiting reagent:
- To find the limiting reagent, compare the moles of sulfuric acid (H2SO4) with the stoichiometric ratio.
- Since the stoichiometric ratio is 2:1, the moles of sulfuric acid (0.316 mol) are greater than twice the moles of sodium hydroxide (2 * 0.2875 mol = 0.575 mol).
- Therefore, sodium hydroxide is the limiting reagent.
7. Calculate the moles of sodium sulfate (Na2SO4) formed from the limiting reagent:
- From the stoichiometry, we know that 1 mole of sodium hydroxide reacts to produce 1 mole of sodium sulfate.
- So, the moles of sodium sulfate formed are equal to the moles of the limiting reagent (0.2875 mol).
8. Calculate the mass of sodium sulfate (Na2SO4) formed:
- Mass = Moles * Molar mass = 0.2875 mol * (142 g/mol) = 40.9 g
Therefore, the maximum mass of sodium sulfate that could be produced by the chemical reaction is approximately 40.9 g (rounded to 3 significant digits).