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Oduvanchick [21]
3 years ago
10

2. How many grams of N2 are required to produce 10.0 grams of NH3? N2 + 3 H2 - 2 NH3

Chemistry
1 answer:
SOVA2 [1]3 years ago
6 0

8.24g

Explanation:

Given parameters;

Mass of NH₃ = 10g

Unknown:

Mass of N₂ = ?

Solution:

  Reaction equation:

               N₂    +    3H₂     →      2NH₃

Step 1: Convert 10.0 g of NH₃ into moles of NH₃

number of moles = \frac{mass}{molar mass}

   molar mass of NH₃ = 14 + 3(1) = 17g/mol

 Number of moles = \frac{10}{17}  = 0.59mole

Step 2: Convert the moles of NH₃ calculated in Step 1 into moles of N₂

       1 mole of N₂    produced 2 mole of NH₃

   

0.59 moles of ammonia would be formed by \frac{0.59}{2} mole = 0.29mole of N₂

Step 3: Convert the moles of N₂ calculated in Step 2 into grams of N₂.

Mass of N₂ = number of moles x molar mass

  Molar mass of N₂ = 14 x 2 = 28g/mol

  Mass of N₂ = 0.29 x 28 = 8.24g

learn more:

Number of moles brainly.com/question/1841136

#learnwithBrainly

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According to the equation below, how many moles of PbO are required to generate 3.88×1023 nitrogen molecules?
saul85 [17]

Answer:

1.935 mole

Explanation:

We'll begin by calculating the number of mole present in 3.88x10^23 molecules of nitrogen(N2). This can be obtained as follow:

From Avogadro's hypothesis, 1 mole of any substance contains 6.02x10^23 molecules. Therefore 1 mole of N2 contains 6.02x10^23 molecules.

Now if 1 mole of N2 contains 6.02x10^23 molecules,

Then Xmol of N2 will contain 3.88x10^23 molecules i.e

Xmol of N2 = (3.88x10^23)/6.02x10^23

Xmol of N2 = 0.645 mole

Now, we can obtain the number of moles of PbO required to generate 3.88x10^23 molecules (i.e 0.645 mole) of N2. This is illustrated below:

The equation for the reaction is given below:

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From the balanced equation above, 3 moles of PbO produced 1 mole of N2.

Therefore, Xmol of PbO will produce 0.645 mole of N2 i.e

Xmol of PbO = 3 x 0.645

Xmol of PbO = 1.935 mole.

From the calculations made above,

1.935 mole of PbO will produce 3.88x10^23 molecules of nitrogen (N2).

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