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Anit [1.1K]
3 years ago
12

Phosphorus reacts with oxygen to form diphosphorus pentoxide: 4 P(s) + 5 O 2(g) -> 2 P 2O 5(s) If 0.97 moles of phosphorus ar

e reacted but 0.12 moles remain after reaction, how many moles of P 2O 5 were produced?
Chemistry
1 answer:
KengaRu [80]3 years ago
6 0

Answer:

0.425 moles of P₂O₅.

Explanation:

As per given balanced equation, four moles of phosphorus reacts with five moles of oxygen to give two moles of P₂O₅.

As given that the initial moles of phosphorus taken = 0.97 moles

moles of phosphorus left after reaction = 0.12 moles

moles of phosphorus reacted = 0.97-0.12 = 0.85 moles

When four moles of P reacts they give two moles of P₂O₅.

when one mole of P will react to give = \frac{2}{4}molP_{2}O_{5}

0.85 moles of P will react to give = \frac{2X0.85}{4}= 0.425molP₂O₅

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Explanation:

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7 0
3 years ago
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250 mL of a solution of calcium oxalate is the evaporated until only a residue of solid calcium
Lana71 [14]

Answer:

2.3 * 10^-5

Explanation:

Recall that the solubility of a solute is the amount of solute that dissolves in 1 dm^3 or 1000cm^3 of solution.

Hence;

Amount of calcium oxalate = 154 * 10^-3/128.097 g/mol = 1.2 * 10^-3 mols

From the question;

1.2 * 10^-3 mols dissolves in 250 mL

x moles dissolves in 1000mL

x = 1.2 * 10^-3 mols * 1000/250

x= 4.8 * 10^-3 moldm^-3

CaC2O4(s) ------->Ca^2+(aq) + C2O4^2-(aq)

Hence Ksp = [Ca^2+] [C2O4^2-]

Where;

[Ca^2+] = [C2O4^2-] = 4.8 * 10^-3 moldm^-3

Ksp = (4.8 * 10^-3)^2

Ksp = 2.3 * 10^-5

4 0
2 years ago
What happens when nitrogen and hydrogen comes in contact at high temperature and pressure in presence of iron powder and molybde
Sidana [21]

Answer:

<h2>Ammonia Gas</h2>

Explanation:

It result in formation of ammonia gas.

N2 + 3H2 ---<u>iron</u><u>/</u><u>molybdenum</u><u>/</u><u>high</u><u> </u>temp/pres--- > 2 NH3

It forms ammonia gas.

Please mark branliest if you are satisfied with the answer. Thanking in anticipation.

6 0
3 years ago
A.it is physical change
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3 years ago
What mass of chromium would be produced from the reaction of 57.0 g of potassium with 199 g of chromium(II) bromide according to
fredd [130]

Answer:

Mass of Chromium produced = 37.91 grams

Explanation:

2K + CrBr₂  →  2KBr + Cr

2mole     1 mole                1 mole

mass of Potassium = 57.0 grams

molar mass of Potassium = 39.1 g/mol

no of moles of Potassium = 57.0 / 39.1 = 1.458 moles

mass of CrBr₂= 199 grams

molar mass of CrBr₂ = 211.8 gram/mole

no of moles of CrBr₂ = 199 / 211.8 = 0.939 mole

From chemical equation

1 mole of CrBr₂ = 2 moles of K

∴ 0.939 moles of CrBr₂ = ?

   ⇒ 0.939 x 2/1 = 1.878 moles of K

1.878 moles of K is needed, but there is 1.458 moles of K. So, Potassium is completed first during the reaction . Hence, Potassium is limiting reagent. and CrBr₂ is excess reagent .

From chemical equation

2 moles of K = 1 mole of Cr

∴ 1.458 moles of K = ?

   ⇒ 1.458 x 1/ 2 = 0.729 moles of Cr

no of moles of Cr formed = 0.729 moles

molar mass of Cr = 52.0 g/mol

mass of one mole of Cr = 52.0 grams

mass of 0.729 moles of Cr = 52.0 x 0.729 = 37.908 grams

mass of Chromium produced = 37.91 grams

6 0
3 years ago
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