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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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P = Pressure in kPa
V = Volume in L
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T = Temperature in Kelvin 

First convert 2.5 atm into kPa:

2.5 X 101.3 = 253.25 kPa

Convert 125 Celsius into Kelvin:

125 + 273 = 398 K

Convert Gallons to Litres:

1.25  X 3.79 = 4.74 L

Plug your values into the equation to solve for n:

(253.25)(4.74) = n(8.314)(398)

n = (253.25)(4.74)/(8.314)(398)

n = 0.362 moles

Now use M = m/n to solve for the mass of O2

M = Molar Mass 
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32 = m/(0.362)

m = (32)(0.362) 

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RUDIKE [14]

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3 years ago
Which describes the earth as an open system with respect to energy?
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Answer:

Explanation:

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6 0
2 years ago
Find the empirical formula of the compound ribose, a simple sugar often used as a nutritional supplement. A 14.229 g sample of r
MakcuM [25]

Answer:

CH2O

Explanation:

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C = 5.692/14.229 * 100 = 40%

O = 7.582/14.229 * 100 = 53.29%

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We then proceed to divide each percentage composition by their atomic mass of 12, 16 and 1 respectively.

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H = 6.71/2 = 6.71

Dividing by the smaller value which is 3.33

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6 0
3 years ago
Read 2 more answers
How many miles of MgCI2 are there in 345 g of the compound?
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MgCl2
Mg=24
Cl=35.5×2 = 71
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345g= 1/95 × 345
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