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Andru [333]
2 years ago
15

2B + 3H2 2BH3 How many moles of BH3 are synthesized when 27.0 moles of H2 completely react?

Chemistry
1 answer:
STatiana [176]2 years ago
3 0

Answer: 18.0 moles of BH_3 will be produced when 27.0 moles of H_2 completely react.

Explanation:

The balanced chemical equation is:

2B+3H_2\rightarrow 2BH_3  

According to stoichiometry :

3 moles of H_2 produce = 2 moles of BH_3

Thus 27.0 moles of H_2 will produce=\frac{2}{3}\times 27.0=18.0 moles  of BH_3

Thus 18.0 moles of BH_3 will be produced when 27.0 moles of H_2 completely react.

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earth has a size of 6,371 km wich is greater than mars size 3,389.5 km

Explanation:

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3 years ago
Based on the reactivities of the elements involved, which reaction will form products that are more stable than the reactants?
MrRa [10]

Answer:

Exothermic reaction

Explanation:

Since in an exothermic reaction heat is released, the products will be more stable than the reactants.

6 0
2 years ago
Wolff-kishner reduction (hydrazine, koh, ethylene glycol, 130°c) of the compound shown gave compound
telo118 [61]
The scheme is shown below, the steps involved are as follow,

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8 0
3 years ago
Explain the difference between an endothermic and an exothermic reaction using the concepts of potential energy, stability, and
kakasveta [241]
An Exothermic reaction releases energy into the surroundings and so the products have more potential energy then the reactants. The enthalpy change is a negative value. Whereas, an endothermic reaction involves the absorption of energy into the system and so the reactants have more potential energy than the products. The enthalpy change is a positive value. This is clearly represented in energy profile diagrams.
4 0
3 years ago
The reaction of hydrogen and iodine to produce hydrogen iodide has a Kc of 54.3 at 703 K. Given the initial concentrations of H2
pentagon [3]

Answer:

[HI] = 0.7126 M

Explanation:

Step 1: Data given

Kc = 54.3

Temperature = 703 K

Initial concentration of H2 and I2 = 0.453 M

Step 2: the balanced equation

H2 + I2 ⇆ 2HI

Step 3: The initial concentration

[H2] = 0.453 M

[I2] = 0.453 M

[HI] = 0 M

Step 4: The concentration at equilibrium

[H2] = 0.453 - X

[I2] = 0.453 - X

[HI] = 2X

Step 5: Calculate Kc

Kc = [Hi]² / [H2][I2]

54.3 = 4x² / (0.453 - X(0.453-X)

X = 0.3563

[H2] = 0.453 - 0.3563 = 0.0967 M

[I2] = 0.453 - 0.3563 = 0.0967 M

[HI] = 2X = 2*0.3563 = 0.7126 M

3 0
3 years ago
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