Answer:
sodium hydroxide is the limiting reactant
Explanation:
The first step is usually to put down the balanced reaction equation. This is the first thing to do when solving any problem related to stoichiometry. The balanced reaction equation serves as a guide during the solution.
2NBr3 + 3NaOH = N2 + 3NaBr + 3HOBr
Let us pick nitrogen gas as our product of interest. Any of the reactants that gives a lower number of moles of nitrogen gas is the limiting reactant.
For nitrogen tribromide
From the balanced reaction equation;
2 moles of nitrogen tribromide yields 1 mole of nitrogen gas
4.3 moles of nitrogen tribromide will yield 4.3 ×1/ 2 = 2.15 moles of nitrogen gas
For sodium hydroxide;
3 moles of sodium hydroxide yields 1 mole of nitrogen gas
5.9 moles of sodium hydroxide yields 5.9 × 1/ 3= 1.97 moles of nitrogen gas
Therefore, sodium hydroxide is the limiting reactant.
<span>all of the above can be saturated molecules </span>
The balanced equation between NaOH and H₂SO₄ is as follows
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
number of moles of NaOH moles reacted = molarity of NaOH x volume
number of NaOH moles = 0.08964 mol/L x 27.86 x 10⁻³ L = 2.497 x 10⁻³ mol
according to molar ratio of 2:1
2 mol of NaOH reacts with 1 mol of H₂SO₄
therefore 2.497 x 10⁻³ mol of NaOH reacts with - 1/2 x 2.497 x 10⁻³ mol of H₂SO₄
number of moles of H₂SO₄ reacted - 1.249 x 10⁻³ mol
Number of H₂SO₄ moles in 34.53 mL - 1.249 x 10⁻³ mol
number of H₂SO₄ moles in 1000 mL - 1.249 x 10⁻³ mol / 34.53 x 10⁻³ L = 0.03617 mol
molarity of H₂SO₄ is 0.03617 M
Carbon(C):
number of moles= mass/molar mass(Mr)
=65.5/12
=5.5 moles
Hydrogen(H):
number of moles=mass/molar mass (Mr)
=5.5/1
=5.5 moles
Oxygen (O):
number of moles = mass/molar mass (Mr)
=29.0/16
=1.8 moles
EF= lowest number of moles over each of the elements
So,
C= 5.5/1.8 = 3
H= 5.5/1.8 = 3
O= 1.8/1.8 = 1
Therefore Emperical formula= C3H3O