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Furkat [3]
3 years ago
13

What is the percent of Cr in Cr2O3

Chemistry
2 answers:
IgorLugansk [536]3 years ago
5 0
The percentage of Chromium in Chromium Oxide is calculated as follow,

Step 1:
           Calculate Molar mass of Cr₂O₃,
Cr = 51.99 u
O = 16 u
So,
     2(51.99) + 3(16) = 103.98 + 48 = 151.98 u

Step 2:
          Secondly divide molar mass of only chromium with total mass of Cr₂O₃ and multiply with 100.
i.e.
                   = \frac{103.98}{151.98} × 100
                   
                   = 68.41 %
So, the %age composition of chromium in chromium oxide is 68.41 %.
lys-0071 [83]3 years ago
4 0

Answer:

68.41

Explanation:

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Answer:

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3 0
3 years ago
How many grams of H2 are needed to react with 2.40 g of N2?
Georgia [21]
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5 0
4 years ago
If you feed 100 kg of N2 gas and 100 kg of H2 gas into a
torisob [31]

Answer : The mass of ammonia produced can be, 121.429 k

Solution : Given,

Mass of N_2 = 100 kg  = 100000 g

Mass of H_2 = 100 kg = 100000 g

Molar mass of N_2 = 28 g/mole

Molar mass of H_2 = 2 g/mole

Molar mass of NH_3 = 17 g/mole

First we have to calculate the moles of N_2 and H_2.

\text{ Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molar mass of }N_2}=\frac{100000g}{28g/mole}=3571.43moles

\text{ Moles of }H_2=\frac{\text{ Mass of }H_2}{\text{ Molar mass of }H_2}=\frac{100000g}{2g/mole}=50000moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

N_2+3H_2\rightarrow 2NH_3

From the balanced reaction we conclude that

As, 1 mole of N_2 react with 3 mole of H_2

So, 3571.43 moles of N_2 react with 3571.43\times 3=10714.29 moles of H_2

From this we conclude that, H_2 is an excess reagent because the given moles are greater than the required moles and N_2 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of NH_3

From the reaction, we conclude that

As, 1 mole of N_2 react to give 2 mole of NH_3

So, 3571.43 moles of N_2 react to give 3571.43\times 2=7142.86 moles of NH_3

Now we have to calculate the mass of NH_3

\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3

\text{ Mass of }NH_3=(7142.86moles)\times (17g/mole)=121428.62g=121.429kg

Therefore, the mass of ammonia produced can be, 121.429 kg

6 0
3 years ago
A sample of potassium chlorate (15.0 g) is dissolved in 201 g of water at 70 °c with precautions taken to avoid evaporation of a
zlopas [31]
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4 0
3 years ago
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