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Y_Kistochka [10]
3 years ago
10

Can someone pls help me

Chemistry
2 answers:
Veronika [31]3 years ago
5 0

Answer:

22(don't forget to add the degrees

Explanation:

2x - 33 + 5x - 31 = 90

2x + 5x = 90 + 31 + 33

7x = 154

x = 154 \div 7

x = 22

Alchen [17]3 years ago
3 0

Answer:

A

Explanation:

That angle is a complementary angle which adds up to 90 degrees. So you write out the equation to solve for X. Equation= 2x-33+5x-31=90. Solve for x and you get 22 degrees.

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

Solution : Given,

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Mass of H_2 = 100 kg = 100000 g

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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

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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.

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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

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