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3 years ago

Someone please help me

Chemistry

1 answer:

ehidna [41]3 years ago

8 0

I hop this will help u

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they belong to same grop

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The method by which the substances in a mixture are separated is called what?

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Answer:

Filtration is a separation method used to separate out pure substances in mixtures comprised of particles some of which are large enough in size to be captured with a porous material.

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3 years ago

What element is found in group 5 period 3?

tatuchka [14]

Answer:

Phosphorus

Explanation:

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3 years ago

A glass jar is sealed, trapping air inside. The jar warms up after it is left ourside in the sun. What will happpen to the press

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3 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