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Alex_Xolod [135]
3 years ago
12

Under appropriate conditions, nitrogen and hydrogen undergo a combination reaction to yield ammonia: N 2 (g) 3H 2 (g) 2NH3 (g) I

f the reaction yield is 87.5%, how many moles of N 2 are needed to produce 3.00 mol of NH 3
Chemistry
1 answer:
balandron [24]3 years ago
5 0

Answer:

We need 1.714 moles N2

Explanation:

Step 1: Data given

The reaction yield = 87.5 %

Number of moles NH3 produced = 3.00 moles

Step 2: The balanced equation

N2(g)+ 3H2(g) →2NH3(g)

Step 3: Calculate moles N2

For 2 moles NH3 produced we need 1 mol N2 and 3 moles H2

This means, if the yield was 100%, for 3.00 moles NH3 produced , we need 1.5 moles N2

For a 87.5 % yield:  

we need more N2, increased by a ratio of 100/87.5.

100/87.5 * 1.5 =  1.714 moles N2

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(will give brainliest) show your work. How many grams of Copper(I) nitrate, CuNO3 are required to produce 88.0 grams of aluminum
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Based on the stoichiometry of the reaction, 156.114 g of CuNO3 are required to produce 88.0 grams of aluminum nitrate, Al(NO3)3.

<h3>What is stoichiometry of a reaction?</h3>

The stoichiometry of a reaction is the molar ratio in which reactants combine to form products.

The stoichiometry of the reaction shows that 6 moles of copper (i) nitrate produces 2 moles of aluminium nitrate.

molar mass of Copper(I) nitrate, CuNO3 = 126 g

molar mass of aluminum nitrate, Al(NO3)3 = 213 g

88.0 g of aluminum nitrate, Al(NO3)3 = 88.0/213 moles = 0.413 moles

0.413 moles of Al(NO3)3 will be produced by 0.413 ×6/3 = 1.239 moles of CuNO3

Mass of 1.239 moles of CuNO3 = 1.239 × 126 = 156.114 g of CuNO3

Therefore, 156.114 g of CuNO3 are required to produce 88.0 grams of aluminum nitrate, Al(NO3)3.

Learn more about stoichiometry at: brainly.com/question/16060223

Therefore, 156.114 g of CuNO3

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                              Moles  =  # of Particles / 6.022 × 10²³

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