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koban [17]
2 years ago
12

How many grams of O2 are reacted when 3.5 mols of P203 are formed?

Chemistry
1 answer:
Neporo4naja [7]2 years ago
7 0
1.45 moles of P2O3 contains 1.45 moles of O3 which means 4.35 moles of O:

1.45 × 3 = 4.35 moles of O

Then you can calculate moles of O2 by dividing it by 2:

4.35 moles ÷ 2 = 2.175 moles of O2 needed
You might be interested in
How many grams are in 2.3 moles of calcium phosphate, Ca3(PO3)2
irina1246 [14]
Moles = mass/molar mass
moles = 2.3
molar mass = 278
=> mass = moles*molar mass = 639.4g
6 0
3 years ago
A student reacts 5.0 g of sodium with 10.0 g of chlorine and collect 5.24 g of sodium chloride. What is the percent yield of thi
Ede4ka [16]

Answer: The percent yield of this combination reaction is 41.3 %

Explanation : Given,

Mass of Na = 5.0 g

Mass of Cl_2 = 10.0 g

Molar mass of Na = 23 g/mol

Molar mass of Cl_2 = 71 g/mol

First we have to calculate the moles of Na and Cl_2.

\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}

\text{Moles of }Na=\frac{5.0g}{23g/mol}=0.217mol

and,

\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}

\text{Moles of }Cl_2=\frac{10.0g}{71g/mol}=0.141mol

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation will be:

2Na+Cl_2\rightarrow 2NaCl

From the balanced reaction we conclude that

As, 2 mole of Na react with 1 mole of Cl_2

So, 0.217 moles of Na react with \frac{0.217}{2}=0.108 moles of Cl_2

From this we conclude that, Cl_2 is an excess reagent because the given moles are greater than the required moles and Na is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of NaCl

From the reaction, we conclude that

As, 2 mole of Na react to give 2 mole of NaCl

So, 0.217 mole of HCl react to give 0.217 mole of NaCl

Now we have to calculate the mass of NaCl

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

Molar mass of NaCl = 58.5 g/mole

\text{ Mass of }NaCl=(0.217moles)\times (58.5g/mole)=12.7g

Now we have to calculate the percent yield of this reaction.

Percent yield = \frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100

Actual yield = 5.24 g

Theoretical yield = 12.7 g

Percent yield = \frac{5.24g}{12.7g}\times 100

Percent yield = 41.3 %

Therefore, the percent yield of this combination reaction is 41.3 %

4 0
3 years ago
PLEASE HELP!
fgiga [73]
150/30 = 5
HF1 20/2 = 10
HF2 10/2 = 5
HF3 5/2 = 2.5
HF4 2.5/2 = 1.25
HF5 1.25/2 = 0.625
Answer: 0.63g
5 0
2 years ago
Read 2 more answers
How many moles of HCl are present in 0.70 L of a 0.63 M HCl solution?
eimsori [14]
So the molarity equation is moles of solute/liters of solution. so i’m pretty sure the answer should be 0.63/0.70= .9
7 0
3 years ago
What is the [H+] of a solution with a pH of 3? Use a scientific calculator.
Mariulka [41]

Answer:

1 x 10⁻³.

Explanation:

∵ pH = - log[H⁺]

3 = - log[H⁺].

log[H⁺] = -3.

∴ [H⁺] = 1 x 10⁻³.

3 0
3 years ago
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