All categories

2 years ago

4 b + 3 o2 → 2 b2o3 if 8 moles of b and 4 moles of o2 are allowed to react, how many moles of b2o3 can be formed?

Chemistry

1 answer:

raketka [301]2 years ago

6 0

The number of moles of b2o3 that will be formed is determined as 4 moles.

<h3>Limiting reagent</h3>

The limiting reagent is the reactant that will be completely used up.

4 b + 3O₂ → 2b₂O₃

from the equation above;

4 b ------------> 2 b₂O₃

2b ------------> b₂O₃

2 : 1

3O₂ -------------> 2b₂O₃

3 : 2

b is the limiting reagent, thus, the amount of b2o3 to be formed is calculated as;

4 b ------------> 2 moles of b2o3

8 moles -------> ?

= (8 x 2)/4

= 4 moles

Thus, the number of moles of b2o3 that will be formed is determined as 4 moles.

Learn more about limiting reactants here: brainly.com/question/14222359

#SPJ1

You might be interested in

Assessment

MakcuM [25]

i didn’t mean to put this opps

4 0

3 years ago

What mass of aluminum is needed to produce 0.500 mole of aluminum chloride?

3241004551 [841]

Answer: " 13.5 g Al " ;

→ that is: "13.5 grams of aluminum."

____________________________

Explanation:

____________________________

Note: What is missing from the question is the "balanced chemical equation" for the "chemical reaction" that contains:

The reactants: "aluminum (Al) " ; and "chlorine (Cl) " ; and:

The product: "aluminum choloride (AlCl₃) " .

____________________________

The "balanced chemical equation" is:

____________________________

2 Al + 3 Cl₂ → 2 AlCl₃ ;

_____________________________

Note: The molecular weight of "aluminum (Al)" is: " 26.98 g /mol " .

____________________________

So: We call solve using a technique known as: "dimensional analysis" :

____________________________

0.500 mol AlCl₃ * $(\frac{2mol Al}{2mol AlCl_{3} }) * (\frac{26.98g Al}{1 mol Al}) = ?$

____________________________

Note: The units of "mol AlCl₃" cancel out to "1' ; and:

The units of "mol Al" cancel out to "1" ; and we are left with:

____________________________

" $\frac{(0.500 * 2 * 26.98)}{2}$ g Al ["grams of aluminum"] ;

____________________________

Note: We can "cancel out the "2's" ; since "2/2 = 1 " ; and we have:

→ (0.500 * 26.98) g Al ;

= 13.49 g Al ;

→ Round to 3 (Three) significant figures;

→ Since: "0.500" has 3 (Three) significant figures:

____________________________

= 13.5 g Al ; that is: "13.5 grams of aluminum."

____________________________

Hope this is helpful!

Best wishes to you in your academic pursuits—and within the "Brainly" community!

____________________________

3 0

3 years ago

PLEASE HELP ASAP

erma4kov [3.2K]

In the so called rain shadow effect we have interaction between all of the four major Earth spheres. When we have a coastal region where there's a high mountain range, the part of the mountain that is facing the sea will differ a lot from the part of the mountain that is on the other side. The water from the sea evaporates. The water vapor makes the air wet. The warm and wet air masses from the sea will come to the coastline, once they reach the mountain they will start to accumulate as they can not pass through it. As they accumulate rainfall appears. The rainfall contributes to a lush vegetation on this side of the mountain (windward side). The rain shadow effect appears on the leeward side of the mountain, and it mostly gets dry, strong, downward winds. These conditions result in drier climate, much less vegetation, and much increased erosion. Thus we can easily see that we have in this case interaction between the hydrosphere (the sea and the rainfall), the geosphere (the ground, soil, rocks), biosphere (the vegetation), and atmosphere (the winds, the clouds).

6 0

4 years ago

If we add a catalyst to the following equation, CO + H2O + heat CO2 + H2, which way will the equilibrium shift?

Igoryamba

Answer:

No effect.

Explanation:

Hello,

In this case, considering the widely studied Le Chatelier's principle, we can realize that the factors affecting equilibrium are concentration, temperature and pressure and volume if the reaction is in gaseous phase and with non-zero change in the number of moles. In such a way, by adding a catalyst to given reaction will have no effect on the equilibrium direction.

Best regards.

7 0

3 years ago

The addition of 250.0 J to 30.0 g of copper initially at 22.0°C will change its temperature to what final value? (Specific heat

WINSTONCH [101]

Answer:

Final temperature = $43.53^{\circ} C$

Explanation:

Given that,

Heat added, Q = 250 J

Mass, m = 30 g

Initial temperature, T₁ = 22°C

The Specific heat of Cu= 0.387 J/g °C

We know that, heat added due to the change in temperature is given by :

$Q=mc\Delta T\\\\Q=mc(T_2-T_1)\\\\\dfrac{Q}{mc}=(T_2-T_1)\\\\T_2=\dfrac{Q}{mc}+T_1$

Put all the values,

$T_2=\dfrac{250}{30\times 0.387}+22\\\\=43.53^{\circ} C$

So, the final temperature is equal to $43.53^{\circ} C$ .

8 0

3 years ago