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frozen [14]
4 years ago
12

elf-contained self-rescue breathing devices convert CO2 into O2 according to the following reaction: 4KO2(s) + 2CO2(g) → 2K2CO3(

s) + 3O2(g) How many grams of KO2 are needed to produce 100.0 L of O2 at 20.0 °C and 1.00 atm?
Chemistry
1 answer:
dmitriy555 [2]4 years ago
7 0

Answer:

394.05g of KO2

Explanation:

First, let us calculate the number of moles of O2 produced. This is illustrated below:

V = 100L

P = 1atm

T = 20°C = 20 +273 = 293K

R = 0.082atm.L/K /mol

n =?

PV = nRT

n = PV /RT

n = (1 x 100)/(0.082 x 293)

n = 4.16moles of O2

4KO2 + 2CO2 → 2K2CO3 + 3O2

From the equation,

4moles of KO2 produced 3moles of O2.

Therefore, xmol of KO2 will produce 4.16moles of O2 i.e

xmol of KO2 = (4.16 x 4)/3 = 5.55moles.

Converting 5.55moles O KO2 to grams, we have:

Molar Mass of KO2 = 39 + (2x16) = 39 + 32 = 71g/mol

Mass of KO2 = 5.55 x 71 = 394.05g

Therefore, 394.05g of KO2 is needed for the reaction

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The mass of sodium chloride in (g) is 14.19 The volume of ammonia solution in (mL) is 36.15 Calculate the following: What is the
Svetach [21]

This is an incomplete question, here is a complete question.

A weighed amount of sodium chloride is completely dissolved in a measured volume of 4.00 M ammonia solution at ice temperature, and carbon dioxide is bubbled in. Assume that sodium bicarbonate is formed util the limiting reagent is entirely used up. the solubility of sodium bicarbonate in water at ice temperature is 0.75 mol per liter. Also assume that all the sodium bicarbonate precipitated is collected and converted quantitatively to sodium carbonate.

Data to be used for calculating the results

-The mass of sodium chloride in (g) is 14.19

-The volume of ammonia solution in (mL) is 36.15

Calculate the following: What is the theoretical yield of sodium bicarbonate in grams?

Answer : The theoretical yield of sodium bicarbonate in grams is, 20.4 grams.

Explanation :

First we have to calculate the moles of NaCl and NH_3.

\text{ Moles of }NaCl=\frac{\text{ Mass of }NaCl}{\text{ Molar mass of }NaCl}=\frac{14.19g}{58.5g/mole}=0.243moles

\text{ Moles of }NH_3=\text{ Concentration of }NH_3\times \text{ Volume of solution}=4.00M\times 0.3615L=1.446moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction will be:

NH_3+NaCl+CO_2+H_2O\rightarrow NaHCO_3+NH_4Cl

From the balanced reaction we conclude that

As, 1 mole of NaCl react with 1 mole of NH_3

So, 0.243 mole of NaCl react with 0.243 mole of NH_3

From this we conclude that, NH_3 is an excess reagent because the given moles are greater than the required moles and NaCl is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of NaHCO_3

From the reaction, we conclude that

As, 1 mole of NaCl react to give 1 mole of NaHCO_3

So, 0.243 moles of NaCl react to give 0.243 moles of NaHCO_3

Now we have to calculate the mass of NaHCO_3

\text{ Mass of }NaHCO_3=\text{ Moles of }NaHCO_3\times \text{ Molar mass of }NaHCO_3

Molar mass of sodium bicarbonate = 84 g/mol

\text{ Mass of }NaHCO_3=(0.243moles)\times (84g/mole)=20.4g

Thus, the theoretical yield of sodium bicarbonate in grams is, 20.4 grams.

4 0
4 years ago
Cadmium bromide is used in photography and lithography. What is the molality of a solution prepared by dissolving 45.38 g of CdB
mash [69]

Answer:

0.44 m is the molality

Explanation:

Molality means mol of solute in 1kg of solvent.

Solute → CdBr₂

Solvent → H₂O

Mass / Molar mass = 45.38 g / 272.21 g/m = 0.167 moles

1 g = 1x10⁻³ kg

375 g = 0.375 kg

0.167 m/0.375 kg = 0.44 m

6 0
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