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suter [353]
3 years ago
9

To ensure that the air in the shuttle remains free of excess CO2, engineers test the air-purification system. they combine 1.000

× 103 g LiOH with 8.80 × 1 02 g CO2. The reaction produces 3.25 × 102 g H2O. What is the limiting reactants in this test reaction? What is the theoretical yield of this test reaction?
Chemistry
2 answers:
Tema [17]3 years ago
7 0

The

balanced reaction is:

2LiOH + CO2 = Li2CO3 + H2O

We

are given the amount of the reactants to be reacted. This will be the starting point of our

calculations. To determine the limiting reactant, we convert the amounts from grams to moles.

1.000 × 10^3 g LiOH  (1 mol / 25.95 g) = 38.54 mol LiOH

8.80 × 10^2 g CO2 ( 1mol / 44.01 g) = 20.00 mol CO2

From the balanced reaction, the molar ratio of the reactants is 2:1. This means that every two mole of lithium hydroxide, 1 mole of carbon dioxide is needed. It is clear that, from the given amounts, CO2 is the limiting reactant.

loris [4]3 years ago
5 0

<u>Answer:</u> The limiting reagent is carbon dioxide and theoretical yield of the reaction is 360 g.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

  • <u>For LiOH:</u>

Given mass of LiOH = 1.00\times 10^3g

Molar mass of LiOH = 23.95 g/mol

Putting values in equation 1, we get:

\text{Moles of LiOH}=\frac{1.00\times 10^3g}{23.95g/mol}=41.75mol

  • <u>For carbon dioxide:</u>

Given mass of carbon dioxide = 8.8\times 10^2g

Molar mass of carbon dioxide = 44 g/mol

Putting values in equation 1, we get:

\text{Moles of carbon dioxide}=\frac{8.8\times 10^2g}{44g/mol}=20mol

  • The chemical equation for the reaction of lithium hydroxide and carbon dioxide gas follows:

2LiOH(s)+CO_2(g)\rightarrow Li_2CO_3(s)+H_2O(l)

By Stoichiometry of the reaction:

1 moles of carbon dioxide reacts with 2 moles of lithium hydroxide.

So, 20 moles of carbon dioxide will react with = \frac{2}{1}\times 20=40mol of lithium hydroxide.

As, given amount of lithium hydroxide is more than the required amount. So, it is considered as an excess reagent.

Thus, carbon dioxide is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of carbon dioxide produces 1 mole of water

So, 20 moles of carbon dioxide will produce = \frac{1}{1}\times 20=20moles of water

  • Now, calculating the theoretical yield of water from equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 20 moles

Putting values in equation 1, we get:

20mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=360g

Hence, the limiting reagent is carbon dioxide and theoretical yield of the reaction is 360 g.

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3 years ago
A 2.26 M solution of KOH is prepared. Calculate the moles and mass of solute present in a 15.2-mL sample of this solution. The m
kenny6666 [7]

Answer:

0.0344 moles and 1.93g.

Explanation:

Molarity is defined as the ratio between moles of a solute (In this case, KOH), and the volume. With molarity and volume we can solve the moles of solute. With moles of solute we can find mass of the solute as follows:

<em>Moles KOH:</em>

15.2mL = 0.0152L * (2.26mol / L) = 0.0344moles

<em>Mass KOH:</em>

0.0344 moles * (56.11g/mol) = 1.93g of KOH

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What is true of the mass and volume of all the floating objects?
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3 0
2 years ago
Exactly 16 mL of a solution A is diluted to 300 mL, resulting in a new solution B that has 0.50 M concentration. If the solution
nordsb [41]

Answer:

8.77g

Explanation:

Step 1:

Data obtained from the question.

This includes the following:

Concentration of A (C1) =.?

Volume of A (V1) = 16 mL

Volume of B (V2) = 300 mL

Concentration of B (C2) = 0.50 M

Molar Mass of NaCl = 58.443 g/mol

Mass of NaCl =.?

Step 2:

Determination of the concentration of A.

Applying the dilution formula:

C1V1 = C2V2

The concentration of A i.e C1 can be obtained as follow:

C1V1 = C2V2

C1 x 16 = 0.5 x 300

Divide both side by 16

C1 = (0.5 x 300) / 16

C1 = 9.375 M

Therefore, the concentration of A is 9.375 M

Step 3:

Determination of the number of mole of NaCl in 9.375 M NaCl solution. This is illustrated below:

Molarity = 9.375 M

Volume = 16 mL = 16/1000 = 0.016 L

Mole of NaCl =?

Molarity = mole /Volume

Mole =Molarity x Volume

Mole of NaCl = 9.375 x 0.016

Mole of NaCl = 0.15 mole

Step 4:

Determination of the mass of NaCl. This is illustrated below:

Mole of NaCl = 0.15 mole

Molar Mass of NaCl = 58.443 g/mol

Mass of NaCl =?

Mass = number of mole x molar Mass

Mass of NaCl = 0.15 x 58.443

Mass of NaCl = 8.77g

Therefore, 8.77g of NaCl is needed to make 1 L of the original solution A.

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