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Ilya [14]
3 years ago
10

Which of the following you is true for a limiting reactant

Chemistry
1 answer:
skad [1K]3 years ago
3 0

Answer:

  • <em><u>C) The limiting reactant has the lowest ratio of moles available / coefficient in the balanced equation.</u></em>

Explanation:

Please, find attached a complete question to determine which of the statements is or are true for a limiting reactant in a chemical equation.

First, remember that the limiting reactant is the substance that is consumed completely while the excess reactant is the substance that does not react completely.

The limiting reactant is found comparing the stoichiometry ratio and the actual ratio between the reactants.

The stoichiometry ratio is found using the coefficientes of the chemical equation.

For illustration, assume the general chemical equation:

         aA+bB\rightarrow cC+dD

The stoichiometric ratio of the reactants is:

          a\text{ }moles\text{ }of\text{ }A/b\text{ }moles\text{ }of\text{ }B

If the ratio of the available moles of substance A to the available moles of  substance B is greater than the stoichiometric ratio, it means that there are more moles of the substance A than what is needed to react with the available moles of substance B, then A will be in excess and B will B the limiting reactant.

If, on the contrary, the ratio of the available moles of substance A to the available moles of  substance B is is less than the stoichiometric ratio, then substance A is less than the necessary to make the all the moles of substance B react, meaning that the substance A will limit the reaction (it will be consumed completely), while the substance B will be in excess.

As for the options:

<em><u>A) The limiting reactant is has the lowest coefficient in a balanced equation.</u></em>

This is false, since it is not the magnitude of the coefficiente what determines the limiting reactant, but the comparison of the ratios.

<u><em>B) The limiting reactant is the reactant for which you have the fewest number of moles.</em></u>

This is false because it is not the number of moles what determines the limiting reactant , but the comparison of the ratios.

<u><em>C) The limiting reactant has the lowest ratio of moles available / coefficient in the balanced equation.</em></u>

This is true as proved below.

The stoichiometric ratio of the reactants is:

          a\text{ }moles\text{ }of\text{ }A/b\text{ }moles\text{ }of\text{ }B

The actual ratio is:

         available\text{ }moles\text{ }of\text{ }A/available\text{ }moles\text{ }of\text{ }B

Assume the first ratio is less than the second (which describes when the substance A is in excess and the limiting reactant is the substance B).

a\text{ }moles\text{ }of\text{ }A/b\text{ }moles\text{ }of\text{ }B

Change the relation to show the ratios of moles available of each substance to the cofficient in the chemical equation:

available\text{ }moles\text{ }of\text{ }B/b\text{ }moles\text{ }of\text{ }B

Then, in the scenary that the limiting reactant is the substance B, the ratio of the left is lower than the ratio of the right, which is the same that limiting reactant has the lowest ratio of moles available / coefficient in the balanced equation.

<em><u>D) The limiting reactant has the lowest ratio of coefficients in the balanced eqution/moles available.</u></em>

<em><u /></em>

This ratio is the inverse of the ratio of the previous statement, thus the relation is inverse, and, since the previous statement was true, this statement is false.

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How many mL of a 1.48 M calcium hydroxide solution are needed to neutralize 36.0 mL of a 1.63 M hydrochloric acid solution
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The volume (in mL) of calcium hydroxide, Ca(OH)₂ needed for the reaction is 19.8 mL

<h3>Balanced equation </h3>

2HCl + Ca(OH)₂ —> CaCl₂ + 2H₂O

From the balanced equation above,

  • The mole ratio of the acid, HCl (nA) = 2
  • The mole ratio of the base, Ca(OH)₂ (nB) = 1

<h3>How to determine the volume of Ca(OH)₂ </h3>
  • Molarity of base, Ca(OH)₂ (Mb) = 1.48 M
  • Volume of acid, HCl (Va) = 36 mL
  • Molarity of acid, HCl (Ma) = 1.63 M
  • Volume of base, Ca(OH)₂ (Vb) =?

MaVa / MbVb = nA / nB

(1.63 × 36) / (1.48 × Vb) = 2

58.68 / (1.48 × Vb) = 2

Cross multiply

2 × 1.48 × Vb = 58.68

2.96 × Vb = 58.68

Divide both side by 2.96

Vb = 58.68 / 2.96

Vb = 19.8 mL

Learn more about titration:

brainly.com/question/14356286

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Determine the pressure in atm exerted by 1 mole of methane placed into a bulb with a volume of 244.6 mL at 25°C. Carry out two c
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The pressure in atm exerted by 1 mole of methane placed into a bulb with a volume of 244.6 mL at 25°C is 101.94atm.

<h3>How to calculate pressure?</h3>

The pressure of an ideal gas can be calculated using the following formula:

PV = nRT

Where;

  • P = pressure
  • V = volume
  • n = number of moles
  • R = gas law constant
  • T = temperature

According to information in this question;

  • T = 25°C = 25 + 273 = 298K
  • V = 244.6mL = 0.24L
  • R = 0.0821 Latm/Kmol

P × 0.24 = 1 × 0.0821 × 298

0.24P = 24.47

P = 24.47/0.24

P = 101.94atm

Therefore, the pressure in atm exerted by 1 mole of methane placed into a bulb with a volume of 244.6 mL at 25°C is 101.94atm.

Learn more about pressure at: brainly.com/question/11464844

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