How many molecules of water are produced from the reaction of 140 g of sulfuric acid and 180 grams of sodium hydroxide?
How many grams of excess reagent will be left over?
How many moles of sodium sulfate will be produced from this reaction?
To answer these questions, we need to use Stoichiometry, which is a method used to calculate quantities of substances involved in chemical reactions.
First, let's write the balanced equation for the reaction between sulfuric acid (H₂SO₄) and sodium hydroxide (NaOH):
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
From the equation, we can see that 1 mole of sulfuric acid reacts with 2 moles of sodium hydroxide to produce 2 moles of water.
1) How many molecules of water are produced from the reaction of 140 g of sulfuric acid and 180 g of sodium hydroxide?
To find the number of molecules of water produced, we need to convert the given masses of sulfuric acid and sodium hydroxide into moles. We'll use the molar mass of each substance to do this.
The molar mass of sulfuric acid (H₂SO₄) is:
2(1.01 g/mol for hydrogen) + 32.07 g/mol (for sulfur) + 4(16.00 g/mol for oxygen) = 98.09 g/mol
The molar mass of sodium hydroxide (NaOH) is:
22.99 g/mol (for sodium) + 16.00 g/mol (for oxygen) + 1.01 g/mol (for hydrogen) = 39.99 g/mol
Now, we can calculate the number of moles for each substance:
Number of moles of sulfuric acid = Mass of sulfuric acid / Molar mass of sulfuric acid
Number of moles of sulfuric acid = 140 g / 98.09 g/mol ≈ 1.428 mol
Number of moles of sodium hydroxide = Mass of sodium hydroxide / Molar mass of sodium hydroxide
Number of moles of sodium hydroxide = 180 g / 39.99 g/mol ≈ 4.500 mol
Since the reaction ratio tells us that 1 mole of sulfuric acid produces 2 moles of water, we can infer that 1.428 mol of sulfuric acid would produce 2 * 1.428 = 2.856 moles of water.
Therefore, from the given reactants, the reaction would produce approximately 2.856 moles of water molecules.
2) How many grams of excess reagent will be left over?
To determine the excess reagent, we need to identify the limiting reagent first. The limiting reagent is the reactant that is entirely consumed in the reaction, thus determining the maximum quantity of product that can be formed.
To find the limiting reagent, we need to compare the moles of sulfuric acid and sodium hydroxide. From our previous calculation, we know that 1 mole of sulfuric acid produces 2 moles of water.
So, if we consider 1.428 mol of sulfuric acid, it would react with half that number of moles of sodium hydroxide (since the mole ratio of sulfuric acid to sodium hydroxide is 1:2).
Thus, the number of moles of sodium hydroxide required is 0.5 * 1.428 = 0.714 mol.
Comparing this with the actual number of moles of sodium hydroxide given (4.500 mol), we can see that sodium hydroxide is in excess.
To determine the grams of the excess reagent left over, we can subtract the number of moles of sodium hydroxide used from the total number of moles of sodium hydroxide given.
Number of moles of sodium hydroxide left over = Total moles of sodium hydroxide given - Moles of sodium hydroxide used
Number of moles of sodium hydroxide left over = 4.500 mol - 0.714 mol = 3.786 mol
Now, we can calculate the mass of sodium hydroxide left over:
Mass of sodium hydroxide left over = Number of moles of sodium hydroxide left over * Molar mass of sodium hydroxide
Mass of sodium hydroxide left over = 3.786 mol * 39.99 g/mol ≈ 151.41 g
Therefore, approximately 151.41 grams of sodium hydroxide will be left over.
3) How many moles of sodium sulfate will be produced from this reaction?
The balanced equation shows that 1 mole of sulfuric acid reacts with 1 mole of sodium sulfate.
From our previous calculation, we know that 1.428 moles of sulfuric acid reacted. Therefore, 1.428 moles of sodium sulfate will be produced.
Hence, approximately 1.428 moles of sodium sulfate will be produced from this reaction.