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nadezda [96]
2 years ago
6

Consider the following generic chemical equation.

Chemistry
1 answer:
makvit [3.9K]2 years ago
5 0

Answer:

\fbox{ A) A  \:  is  \:  limiting \: reactant }

\fbox{ B) B  \:  is  \:  limiting \: reactant }

\fbox{ C) A  \:  is  \:  limiting \: reactant }

\fbox{ D) A  \:  is  \:  limiting \: reactant }

Explanation:

<em>Given equation:</em>

<em>A+3B \rightarrow \: C</em>

<em>To </em><em>find:</em>

Limiting reactant for corresponding number of moles=?

<em>Solution:</em>

We know that the liming reactant is any atom, ion or molecule which is completely consumed during a reaction and other reactant is still left in reactant vessel.

For the given reaction A+3B→C

for every one mole of A three moles of B are required.

﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌

in A) there is one mole of A and 4 mole of B,

if 1 mole of A will react with 3 moles of B, 1 mole of B will be still there in reaction, the reactant was completely consumed is A which is limiting reactant.

﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌

in B) there is 2 mole of A and 3 mole of B,

if 1 mole of A will react with 3 moles of B, 1 mole of A will be still there in reaction, the reactant was completely consumed is B which is limiting reactant.

﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌

in C) there is 0.5 mole of A and 1.6 mole of B

if one mole of A requires three moles of B to complete the reaction then,

0.5 moles of A will require 1.5 moles of B

if 0.5 mole of A will react with 1.6 moles of B, 0.1 mole of B will be still there in reaction, the reactant was completely consumed is A which is limiting reactant.

﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌

in D) there is 24 mole of A and 72 mole of B

if one mole of A requires three moles of B to complete the reaction then,

24 moles of A will require 72 moles of B.

if 24 mole of A will react with 75 moles of B, 3 mole of B will be left over in reaction, the reactant was completely consumed is A which is limiting reactant.

﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌

\fbox{ A) A  \:  is  \:  limiting \: reactant }

\fbox{ B) B \:  is \:  limiting \:  reactant }

\fbox{ C) A  \:  is  \:  limiting \: reactant }

\fbox{ D) A  \:  is  \:  limiting \: reactant }

<em><u>Thanks for joining brainly community.</u></em>

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Tp_1= Total pressure when only oxygen gas was present.

Final total pressure when 4 moles of helium gas were added:

X'_{O_2}=\frac{4}{8}=\farc{1}{2},X_{He}=\frac{4}{8}=\frac{1}{2}

partial pressure of oxygen in the mixture :

Since, the number of moles of oxygen remains the same, the partial pressure of oxygen will also remain the same in the mixture.

p_{O_2}=Tp_2\times X'_{O_2}=Tp_2\times \frac{1}{2}

Tp_2= Total pressure of the mixture.

from (1)

Tp_1=Tp_2\times X'_{O_2}=Tp_2\times \frac{1}{2}

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Initial pressure (P₁) = 2 atm

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Final temperature (T₂) = 0 °C

Final pressure (P₂) = 1 atm

Final volume (V₂) =?

Next, we shall convert celsius temperature to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T₁) = 50 °C

Initial temperature (T₁) = 50 °C + 273

Initial temperature (T₁) = 323 K

Final temperature (T₂) = 0 °C

Final temperature (T₂) = 0 °C + 273

Final temperature (T₂) = 273 K

Finally, we shall determine the new volume. This can be obtained as follow:

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Initial volume (V₁) = 5 L

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Final pressure (P₂) = 1 atm

Final volume (V₂) =?

P₁V₁ / T₁ = P₂V₂ / T₂

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