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stira [4]
3 years ago
5

How many moles of aluminum oxide will be produced from 0.50 mol of oxygen? 4 al + 3 o2 2 al2o3?

Chemistry
2 answers:
LuckyWell [14K]3 years ago
5 0
From the equation;
 4 Al + 3 O2 = 2 Al2O3
The mole ratio of Oxygen is to Aluminium hydroxide is  3:2.
Therefore; moles of Al2O3 is 
 (0.5/3 )× 2 = 0.333 moles
Therefore; The moles of aluminium oxide will be 0.333 moles
Phoenix [80]3 years ago
4 0

Answer : The number of moles of aluminum oxide produced will be 0.33 moles.

Explanation : Given,

Moles of O_2 = 0.50 mole

The given chemical reaction is:

4Al+3O_2\rightarrow 2Al_2O_3

From the balanced chemical reaction we conclude that,

As, 3 moles of O_2 react to give 2 moles of Al_2O_3

So, 0.50 moles of O_2 react to give \frac{0.50}{3}\times 2=0.33 moles of Al_2O_3

Therefore, the number of moles of aluminum oxide produced will be 0.33 moles.

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6 0
3 years ago
a ridged 2.40L sealed vessel contains He and C3H6 gases. The partial pressure of He is 1.4 atm and that of C3H6 is 1.7 atm, at 5
balandron [24]

Answer:

Partial pressure He → 0.96 atm

Partial pressure C₃H₆  → 1.18 atm

Explanation:

We apply the Charles Gay Lussac law, to solve this. Pressure varies directly proportional to absolute T°, when the volume keeps on constant.

P₁ / T₁ = P₂ / T₂

We convert the T° to absolute T°

55°C + 273 = 328K

-45°C + 273 = 228K

Total pressure = Sum of partial pressures

1.7 atm + 1.4 atm = 3.1 atm

When we apply the formula we would know the new total pressure

3.1 atm / 328K = P₂ / 228K

(3.1 atm / 328K) . 228K = 2.15 atm

As the moles has not been modified with the change of T°, we assume the mole fraction is still the same.

Mole fraction He = Partial pressure He / Total pressure

1.4 atm / 3.1 atm = 0.45

Mole fraction C₃H₆ = Partial pressure C₃H₆/ Total pressure

1.7 atm / 3.1 atm = 0.55

0.45 = Partial pressure He / 2.15 atm

Partial pressure = 0.45 . 2.15 atm → 0.96 atm

0.55 = Partial pressure C₃H₆ / 2.15 atm

Partial pressure = 0.55 . 2.15 atm → 1.18 atm

3 0
3 years ago
Which of the possible compounds has a mass of 163 grams when
Simora [160]

Answer:

CH4

Explanation:

In solving this problem, we must remember that one mole of a compound contains Avogadro's number of elementary entities. These elementary entities include atoms, molecules, ions etc. Recall that one mole of a substance is the amount of substance that contains the same number of elementary entities as 12g of carbon-12. The Avogadro's number is 6.02 × 10^23.

Hence we can now say;

If 163 g of the compound contains 6.13 ×10^24 molecules

x g will contain 6.02 × 10^23 molecules

x= 163 × 6.02 × 10^23 / 6.13 × 10^24

x= 981.26 × 10^23/ 6.13 ×10^24

x= 160.1 × 10^-1 g

x= 16.01 g

x= 16 g(approximately)

16 g is the molecular mass of methane hence x must be methane (CH4)

6 0
3 years ago
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