3 years ago

How many liters of oxygen are needed to react completely with 81.0 g of aluminum at STP?

2 answers:

4 0

I think is C
Al= 108g have Vo2= 67.2L
Al= 81.0g have Vo2= ?
What you need to do is:
67.2 multiply with 81.0 and divided by 108 you will see the answer which is 50.4 L

Gelneren [198K]3 years ago

3 0

Answer:

Volume O₂ at STP = 50.4 Liters

Explanation:

4Al(s) + 3O₂(g) => 2Al₂O₃(s) at STP conditions

81g Al(s) = 81g/27g/mole = 3mole Al

moles O₂ consumed = 4/3(3)moles O₂ = 2.25 moles O₂ consumed

Volume O₂ at STP = 2.25moles x 22.4L/mole = 50.4 Liters O₂

You might be interested in

If water and oil are combine If water and oil are combined in a container, the resulting liquid is a(n) A. solution. B. mixture.

mart [117]

B. Mixture is the answer

8 0

3 years ago

The decomposition of NI3 to form N2 and I2 releases −290.0 kJ of energy. The reaction can be represented as 2NI3(s)→N2(g)+3I2(g)

EastWind [94]

Answer:

-7.34 kilo Joules is the change in enthaply when 20.0 grams of nitrogen triiodide decomposes.

Explanation:

Mass of nitrogen triiodide = 20.0 g

Moles of nitrogen triiodide = $\frac{20.0 g}{395 g/mol}=0.05063 mol$

$2NI_3(s)\rightarrow N_2(g)+3I_2(g), \Delta H_{rxn}=-290.0 kJ$

According to reaction, 2 moles of nitrogen triiodide gives 290.0 kilo Joules of heat on decomposition ,then 0.05063 moles of nitrogen triiodide will give :

$\frac{-290.0 kJ}{2}\times 0.05063=-7.34 kJ$