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2 years ago

Limiting Reactant Worksheet

Chemistry

1 answer:

faltersainse [42]2 years ago

6 0

Answer:

a) 1.61 mol

b) Al is limiting reactant

c) HBr is in excess

Explanation:

Given data:

Moles of Al = 3.22 mol

Moles of HBr = 4.96 mol

Moles of H₂ formed = ?

What is limiting reactant =

What is excess reactant = ?

Solution:

Chemical equation:

2Al + 2HBr → 2AlBr + H₂

Now we will compare the moles:

Al : H₂

2 : 1

3.22 : 1/2×3.22 = 1.61 mol

HBr : H₂

2 : 1

4.96 : 1/2×4.96 = 2.48 mol

The number of moles of H₂ produced by Al are less it will be limiting reactant while HBr is present in excess.

Moles of H₂ :

Number of moles of H₂ = 1.61 mol

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Wewaii [24]

8.50 moles is equal to 5.1187×10²⁴ atoms of Ca.

<u>Explanation:</u>

We have to multiply the moles of Ca by the Avogadro's number:

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So the number of atoms:

= 8.5 moles × 6.022×10²³atoms / mol

= 5.1187×10²⁴ atoms

Hence the 8.50 moles is equal to 5.1187×10²⁴ atoms of Ca.

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Answer:

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Explanation:

Given data:

Mass of Nitrogen = 38.5 g

Moles of ammonia produced = ?

Solution:

Chemical equation:

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2 years ago

Determine the value of the equilibrium constant, KgoalKgoalK_goal, for the reaction CO2(g)⇌C(s)+O2(g)CO2(g)⇌C(s)+O2(g), Kgoal=?

NISA [10]

Answer:

The value of the equilibrium constant for reaction asked is $1.24\times 10^{-7}$ .

Explanation:

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$K_{goal}=\frac{[C][O_2]}{[CO_2]}$

$2CO_2(g)+2H_2O(l)\rightleftharpoons CH_3COOH(l)+2O_2(g)$ ..[1]

$K_1=\frac{[CH_3COOH][O_2]^2}{[CO_2]^2[H_2O]^2}$

$2H_2(g)+O2(g)\rightleftharpoons 2H_2O(l)$ ..[2]

$K_2=\frac{[H_2O]^2}{[H_2]^2[O_2]}$

$CH_3COOH(l)\rightleftharpoons 2C(s)+2H_2(g)+O_2(g)$ ..[3]

$K_3=\frac{[C]^2[H_2]^2[O_2]}{[CH_3COOH]}$

[1] + [2] + [3]

$2CO_2(g)\rightleftharpoons 2C(s)+2O_2(g)$

( on adding the equilibrium constant will get multiplied with each other)

$K=K_1\times K_2\times K_3$

$K=5.40\times 10^{-16}\times 1.06\times 10^{10}\times 2.68\times 10^{-9}$

$K=1.53\times 10^{-14}$

$K=\frac{[C]^2[O_2]^2}{[CO_2]^2}$

On comparing the K and $K_{goal}$ :

$K^2=K_{goal}$

$K_{goal}=\sqrt{K}=\sqrt{1.53\times 10^{-14}}=1.24\times 10^{-7}$

The value of the equilibrium constant for reaction asked is $1.24\times 10^{-7}$ .

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3 years ago