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3 years ago

In the combustion reaction CH4+2O2 yield CO2 + 2H2Owhich reactant has the greatest rate of disappearance?

Chemistry

1 answer:

Maksim231197 [3]3 years ago

6 0

Answer:

To consume the 2.8 moles of CH4 we need 5.6 moles of O2 since the molar ratio is 1:2. We have only 3 moles of O2 ; therefore, O2 is the limiting reactant.

Explanation:

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7. Calculate the amount of energy required heat 100.g to H2O(s) changes to H2O(l) at 0°C

Brrunno [24]

Answer:

33300J

Explanation:

Given parameters:

Mass of ice = 100g

Unknown:

Amount of energy = ?

Solution:

This is a phase change process from solid to liquid. In this case, the latent heat of melting of ice is 3.33 x 10⁵ J/kg.

So;

H = mL

m is the mass

L is the latent heat of melting ice

Now, insert the parameters and solve;

H = mL

mass from gram to kilogram;

100g gives 0.1kg

H = 0.1 x 3.33 x 10⁵ = 33300J

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3 years ago

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9. The picture that would best represent a rock is the image C. The solid molecules are so pact that they have no room for movement.

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3 years ago

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Nataliya [291]

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3 years ago

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4 years ago

450g of chromium(iii) sulfate reacts with excess potassium phosphate. How many grams of potassium sulfate will be produced? (ANS

Irina18 [472]

Answer:

600 g K₂SO₄

Explanation:

First write down the complete, balanced chemical equation for such question. In this case:

Cr₂(SO₄)₃ + 2K₃PO₄ → 3K₂SO₄ + 2CrPO₄

Next, calculate molar masses of required compounds mentioned in the question. In this case it is for Cr₂(SO₄)₃ and K₂SO₄.

Molar mass(MM) of Cr₂(SO₄)₃:

= 2*(MM of Cr) + 3*( MM of S) + 3*4*( MM of O)

= 2*(52) + 3*(32) + 12*(16)

= 104 + 96 + 192

= 392 g

Molar mass(MM) of K₂SO₄:

= 2*(MM of K) + 1*( MM of S) + 4*( MM of O)

= 2*(39) + 32 + 4*(16)

= 78 + 32 + 64

= 174 g

Here comes the concept of Limiting reagent:

The limiting reagent in a chemical reaction is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent, since the reaction cannot continue without it. The other reactants present with this are usually in excess and called excess reactants. If quantities of both the reactants are given, then one should apply unitary method and find out the limiting reagent out of the two. Then, determine the amount of product formed or percentage yield.

Also, 1 mole( 392 g) of Cr₂(SO₄)₃ gives 3 moles( 174*3 = 522 g) of K₂SO₄.

Using unitary method, if 392g of Cr₂(SO₄)₃ gives 522 g of K₂SO₄ , then 450 g of Cr₂(SO₄)₃ will give how much of K₂SO₄?

Yeild of K₂SO₄ : $\frac{522 * 450}{392}$

That is 599.3 g.

Since we have not considered molecular masses of individual atoms to 6 decimal places, this number can be approximated to 600g.

Therefore, 600g of K₂SO₄ is produced.

6 0

3 years ago