A. Moles of O2, needed to produce 7 moles of product?
1 answer:
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Answer:
Volume of = 15.31 mL
Volume of = 4.69 mL
Explanation:
Given that:
the density of the mixture = 1.82 g/mL
From the density of the pure samples
The density of = 1.492 g/mL
The density of = 2.890 g/mL
The total volume of the liquid mixture = 20.0 mL
Suppose the volume of = P ml
and the volume of = Q ml
the sum of their volumes should be equal to the total volume of the mixture
----- (1)
However, we know that Density = mass/volume
∴ mass = density × volume
The equation can now be expressed as:
1.492 g/mL × P mL + 2.890 g/mL × Q mL = 1.82 g/mL × 20 mL ---- (2)
From equation (1) ;
let Q = 20 - P
The replace the value of P into equation (2)
1.492 g/mL × P mL + 2.890g/mL × (20 - P) mL = 1.82 g/mL × 20 mL
1.492 P g + 57.8g - 2.890 P g = 36.4g
1.492 P g - 2.890 P g = 36.4g - 57.8g
-1.398 P g = -21.4g
P = -21.4g/-1.398g
P = 15.31 mL
Q = 20 - P
Q = (20 - 15.31) mL
Q = 4.69 mL
∴
Volume of = 15.31 mL
Volume of = 4.69 mL
Answer:
1,812 wt%
Explanation:
The reactions for this titration are:
2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻
I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻
The moles in the end point of S₂O₃⁻ are:
0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:
4,463x10⁻⁴ moles of S₂O₃⁻× = 2,2315x10⁻⁴ moles of I₃⁻
As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:
2,2315x10⁻⁴ moles of I₃⁻× = 4,463x10⁻⁴ moles of Ce(IV). These moles are:
4,463x10⁻⁴ moles of Ce(IV)× = <em>0,0625 g of Ce(IV)</em>
As the sample has a 3,452g, the weight percent is:
0,0625g of Ce(IV) / 3,452g × 100 = <em>1,812 wt%</em>
I hope it helps!