Answer:
514.5 g.
Explanation:
- The balanced equation of the reaction is: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O.
- It is clear that every 2.0 moles of NaOH react with 1.0 mole of H₂SO₄ to produce 1.0 mole of Na₂SO₄ and 2.0 moles of 2H₂O.
- Since NaOH is in excess, so H₂SO₄ is the limiting reactant.
- We need to calculate the no. of moles of 355.0 g of H₂SO₄:
n of H₂SO₄ = mass/molar mass = (355.0 g)/(98.0 g/mol) = 3.622 mol.
Using cross multiplication:
∵ 1.0 mol H₂SO₄ produces → 1.0 mol of Na₂SO₄.
∴ 3.622 mol H₂SO₄ produces → 3.662 mol of Na₂SO₄.
- Now, we can get the theoretical mass of Na₂SO₄:
∴ mass of Na₂SO₄ = no. of moles x molar mass = (3.662 mol)(142.04 g/mol) = 514.5 g.
<u>Answer:</u> 3 moles of water are formed. The limiting reagent is Oxygen atom and the excess reagent is Hydrogen atom.
<u>Explanation:</u>
From the image provided, there are 8 moles of Hydrogen atom and 3 moles of Oxygen atom.
The equation for the formation of water is given by:
By Stoichiometry,
1 mole of Oxygen atom reacts with 2 moles of Hydrogen atom.
So, 3 moles of Oxygen atom will react with = of Hydrogen atom.
The required amount of Hydrogen atom is less than the given amount of Hydrogen atom, hence, it is considered as an excess reagent.
Therefore, Oxygen atom is the limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
1 moles of oxygen atom produces 1 mole of water molecule.
So, 3 moles of oxygen atom will produce = of water molecule.
Thus, 3 moles of water are formed. The limiting reagent is Oxygen atom and the excess reagent is Hydrogen atom.
Answer:
Explanation:The ether functional group consists of an oxygen atom that forms single bonds with two carbon atoms
Answer:
58g
Explanation:
In order to solve this problem, you must take a look at the solubility graph for potassium nitrate.
Now, the solubility graph shows you how much solute can be dissolved per 100g of water in order to make an unsaturated, a saturated, or a supersaturated solution.
You're looking to make a saturated potassium nitrate solution using
50g of water at 60∘C. Your starting point will be to determine how much potassium nitrate can be dissolved in 100g of water at that temperature in order to have a saturated solution.
As you can see, the curve itself represents saturation.
If you draw a vertical line that corresponds to 60∘C and extend it until it intersects the curve, then draw a horizontal line that connects to the vertical axis, you will find that potassium has a solubility of about
115g per 100g of water. Your answer is 58g of potassium nitrate
<span>Made up of particles packed relatively close together, having an indefinite shape but a definite volume</span>