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iogann1982 [59]
3 years ago
11

a) A reaction container holds 5.33 g of P4 and 3.77 g of O2 and reaction A occurs. If enough oxygen is available then the P4O6 r

eacts further to undergo reaction B. What is the limiting reactant for the formation of P4O6? b) What mass of P4O6 is produced (theoretical yield in grams)? c) If 7.12g of P4O6 were obtained what is the percent yield? d) Will reaction B occur? Why or why not? What mass of excess reactant is left in the reaction container?
Chemistry
1 answer:
Aliun [14]3 years ago
4 0

The given question is incomplete, the complete question is:

Balance the following chemical equations: A) P4+ O2 → P406 B) _P406+ LO2 → P4010 a) A reaction container holds 5.33 g of P4 and 3.77 g of O2 and reaction A occurs. If enough oxygen is available then the P406 reacts further to undergo reaction B. What is the limiting reactant for the formation of P406? b) What mass of P406 is produced (theoretical yield in grams)? c) If 7.12g of P406 were obtained what is the percent yield? d) Will reaction B occur? Why or why not? What mass of excess reactant is left in the reaction container?

Answer:

The balanced reaction will be,

A) P₄ + 3O₂ ⇒ P₄O₆

B) P₄O₆ + 2O₂ ⇒ P₄O₆

a) Based on the given information, the reaction container holds 5.33 grams of P₄ and 3.77 grams of oxygen. Thus, the moles of P₄ will be,

Moles = mass of P₄/Molar mass of P₄ = 5.33 grams/124 g/mole = 0.043 mole

Now the moles of O₂ will be,

Moles = mass of O₂/Molar mass of O₂ = 3.77 grams/32 g/mol = 0.112 mole

Now the moles of P₄O₆ formed when 0.043 moles of P₄ react completely will be = 1/1 × 0.043 = 0.043 mole of P₄O₆

Similarly, the moles of P₄O₆ formed, when 0.112 moles of O₂ react completely will be = 1/3 × 0.112 = 0.0373 mole of P₄O₆

Thus, from the analysis, the maximum moles of P₄O₆ formed will be 0.0373 moles. Therefore, oxygen will be the limiting reagent, which will react completely in the reaction.

b) From the above findings, the maximum moles of P₄O₆ produced is 0.0373 mole. Thus, the theoretical yield of P₄O₆ produced will be,

= Moles of P₄O₆ × Molar mass of P₄O₆

Theoretical yield = 0.0373 mole × 220 g/mole = 8.206 grams

c) Based on the given information, the actual mass of P₄O₆ produced is 7.12 grams.

Hence, percent yield = Actual yield/Theoretical yield * 100

= 7.12/8.206 × 100 = 86.77 %

d) In the given case, reaction B will not take place. This is due to the fact that oxygen is not left for reaction B, which was the limiting regent for reaction A. Here P₄ is the excess reactant, which was left in the reaction.

The initial moles of P₄ is 0.043, O₂ is 0.112, and P₄O₆ is O. The final moles of P₄ is 0.043 -1/3 × 0.112 = 0.0057 mole, O₂ is 0, and P₄O₆ is 0.0373 mole.

Thus, moles of P₄ left is 0.0057 mole. Hence, the mass of P₄ left will be,

= 0.0057 mole × Molar mass of P₄

= 0.0057 mole × 124 g/mole = 0.7068 grams.

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You purchase a bottle of concentrated sulfuric acid from a chemical supplier. The bottle reads Sulfuric acid (95% w/w) plastic c
My name is Ann [436]

Answer:

The molarity of the sulfuric acid is 0.018 M

Explanation:

The molarity of a solution is the number of moles of the solute (sulfuric acid in this case) in a 1-liter solution.

Every 100 g of the solution, we have 95 g sulfuric acid because its concentration is 95% w/w.

With the density, we can calculate how many liters are 100 g of solution:

density = mass / volume

1.85 g / ml = 100 g / volume

volume = 100 g / 1.85 g/ml

volume = 54.1 ml or 0.0541 l

Now, we know that we have 95 g sulfuric acid in 0.0541 l solution. In 1 l, we have then:

1 l * 95g / 0.0541 l = 1.756 g sulfuric acid.

But we want to know how many moles sulfuric acid we have per liter. Then, using the molar mass, we can calculate how many moles there are in 1.756 g sulfuric acid:

1.756 g * 1 mol / 98.08 g = 0.018 mol

The molarity is 0.018 M

4 0
3 years ago
1. This is very important in preventing the spread of pathogens in the kitchen.
maks197457 [2]
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6 0
2 years ago
What is one way to test whether an unknown solution is acidic or basic?
sertanlavr [38]

Answer: There are several ways. The first that comes to mind is a pH meter. A pH electrode Is lowered into the solution, and (Assuming) the pH Meter has been properly calibrated, and the temperature of the solution is set to the calibration of the Meter, the pH can be read directly from an analogue scale or digital readout. Below 7 is acidic, 7 is Neutral, (like Pure Water), and over 7 is Alkaline, or Basic.

A useful, but less accurate method is the use of any number of “pH Indicator Solutions”, which are essentially a type of various colored dyes that change color within differing pH ranges. Usually, if the pH is unknown, a small amount of solution is removed from the container and tested separately - in a “well plate”, or similar method.

These types of dyes, or Indicator Solutions, can be dried upon strips of “pH indicator Paper”, which, depending upon the type can be very useful when carrying out more precisely arrived at pH tests like Titration.

Just to see if a solution is “Acid” or “Base”, Litmus paper is used; “a Red color shows Acidity, and a Blue color, a Base”; ergo, “An Acid Solution will turn Litmus Paper, Red”.

4 0
3 years ago
A solution contains 0.0440 M Ca2 and 0.0940 M Ag. If solid Na3PO4 is added to this mixture, which of the phosphate species would
Olenka [21]

Answer:

C. Ca_3(PO_4)_2  will precipitate out first

the percentage of Ca^{2+}remaining =  12.86%

Explanation:

Given that:

A solution contains:

[Ca^{2+}] = 0.0440 \ M

[Ag^+] = 0.0940 \ M

From the list of options , Let find the dissociation of Ag_3PO_4

Ag_3PO_4 \to Ag^{3+} + PO_4^{3-}

where;

Solubility product constant Ksp of Ag_3PO_4 is 8.89 \times 10^{-17}

Thus;

Ksp = [Ag^+]^3[PO_4^{3-}]

replacing the known values in order to determine the unknown ; we have :

8.89 \times 10 ^{-17}  = (0.0940)^3[PO_4^{3-}]

\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3}  = [PO_4^{3-}]

[PO_4^{3-}] =\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3}

[PO_4^{3-}] =1.07 \times 10^{-13}

The dissociation  of Ca_3(PO_4)_2

The solubility product constant of Ca_3(PO_4)_2  is 2.07 \times 10^{-32}

The dissociation of Ca_3(PO_4)_2   is :

Ca_3(PO_4)_2 \to 3Ca^{2+} + 2 PO_{4}^{3-}

Thus;

Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2

2.07 \times 10^{-33} = (0.0440)^3  [PO_4^{3-}]^2

\dfrac{2.07 \times 10^{-33} }{(0.0440)^3}=   [PO_4^{3-}]^2

[PO_4^{3-}]^2 = \dfrac{2.07 \times 10^{-33} }{(0.0440)^3}

[PO_4^{3-}]^2 = 2.43 \times 10^{-29}

[PO_4^{3-}] = \sqrt{2.43 \times 10^{-29}

[PO_4^{3-}] =4.93 \times 10^{-15}

Thus; the phosphate anion needed for precipitation is smaller i.e 4.93 \times 10^{-15} in Ca_3(PO_4)_2 than  in  Ag_3PO_4  1.07 \times 10^{-13}

Therefore:

Ca_3(PO_4)_2  will precipitate out first

To determine the concentration of [Ca^+] when  the second cation starts to precipitate ; we have :

Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2

2.07 \times 10^{-33}  = [Ca^{2+}]^3 (1.07 \times 10^{-13})^2

[Ca^{2+}]^3 =  \dfrac{2.07 \times 10^{-33} }{(1.07 \times 10^{-13})^2}

[Ca^{2+}]^3 =1.808 \times 10^{-7}

[Ca^{2+}] =\sqrt[3]{1.808 \times 10^{-7}}

[Ca^{2+}] =0.00566

This implies that when the second  cation starts to precipitate ; the  concentration of [Ca^{2+}] in the solution is  0.00566

Therefore;

the percentage of Ca^{2+}  remaining = concentration remaining/initial concentration × 100%

the percentage of Ca^{2+} remaining = 0.00566/0.0440  × 100%

the percentage of Ca^{2+} remaining = 0.1286 × 100%

the percentage of Ca^{2+}remaining =  12.86%

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Answer:

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Explanation:

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