Balance Chemical Equation for combustion of Propane is as follow,
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
According to equation,
1 mole of C₃H₈ on combustion gives = 4 moles of H₂O
So,
5 moles of C₃H₈ on combustion will give = X moles of H₂O
Solving for X,
X = (5 mol × 4 mol) ÷ 1 mole
X = 20 moles of H₂O
Calculating number of molecules for 20 moles of H₂O,
As,
1 mole of H₂O contains = 6.022 × 10²³ molecules
So,
20 moles of H₂O will contain = X molecules
Solving for X,
X = (20 mole × 6.022 × 10²³ molecules) ÷ 1 mol
X = 1.20 ×10²⁵ Molecules of H₂O
We are told that KOH is being used to completely neutral H₂SO₄ according to the following reaction:
KOH + H₂SO₄ → H₂O + KHSO₄
If KOH can completely neutralize H₂SO₄, then there must be an equal amount of moles of each as they are in a 1:1 ratio:
0.025 L x 0.150 mol/L = .00375 mol KOH
0.00375 mol KOH x 1 mole H₂SO₄/1 mole KOH = 0.00375 mol H₂SO₄
We are told we have 15 mL of H₂SO₄ initially, so now we can find the original concentration:
0.00375 mol / 0.015 L = 0.25 mol/L
The concentration of H₂SO₄ being neutralized is 0.25 M.
Answer:- Mole ratio of D to A is 4:3.
Explanations:- Mole ratio for a chemical reaction is the ratio of the coefficients.
The given generic chemical reaction is:
The numbers written in front of each chemical species in the chemical reaction are their moles. For the given generic chemical reaction the coefficient of A is 3 and that of B is 1. So, the mole ratio of A to B is 3:1.
Similarly if we want to write the mole ratio of C to D then it is 1:4.
We are asked to write the mole ratio of D to A. So, like the other ratios, the mole ratio of D to A is 4:3 as the coefficient of D is 4 and A is 3.
The answer would actually be false. I just took the test.
