Answer:
51.2g of CO2
Explanation:
The first step is to balance the reaction equation as shown in the solution attached. Without balancing the reaction equation, one can never obtain the correct answer! Then obtain the masses of octane reacted and carbon dioxide produced from the stoichiometric equation. After that, we now compare it with what is given as shown in the image attached.
The relative molecular mass of acid A : 50 g/mol
<h3>Further explanation</h3>Given
40.0 cm³(40 ml) of 0.2M sodium hydroxide
0.2g of a dibasic acid
Required
the relative molecular mass of acid A
Solution
Titration formula
M₁V₁n₁=M₂V₂n₂
n=acid/base valence(number of H⁺/OH⁻)
NaOH ⇒ n = 1
Dibasic acid = diprotic acid (such as H₂SO₄)⇒ n = 2
mol = M x V
Input the value in the formula :(1 = NaOH, 2=dibasic acid)
0.2 x 40 x 1 = M₂ x V₂ x 2
M₂ x V₂ = 4 mlmol = 4.10⁻³ mol ⇒ mol of Acid A
The relative molecular mass of acid A (M) :
Answer:
44.28 grams.
Explanation:
Let us write the balanced reaction:
As per balanced equation, six moles of fluorine gas will give four moles of PF₃.
The mass of PF₃ required = 120 g
The molar mass of PF₃ = 88g/mol
Moles of PF₃ required =
The moles of fluorine gas required =
the mass of fluorine gas required = moles X molar mass = 0.91x38 = 34.58g
Now this much mass will be required if the reaction is of 100% yield
But as given that the yield of reaction is only 78.1%
The mass of fluorine required =