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lara [203]
3 years ago
9

45 grams of ammonia, NH3, in 0.75 L of solution.

Chemistry
1 answer:
GrogVix [38]3 years ago
4 0
2.3 moles of sodium chloride in 0.45 liters of solution . necessarile correct ... 0.025 L. (1.3 SF). 45 grams of ammonia in 0.75 L of solution. 45g x 1 mol = 2. demol ... anol x 1,50 = 3 mol NaOH x 40,019 = 120g Naolt.
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Answer:

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b) 61.2 g of NH3

c) 4.15 g of H2

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Explanation:

Equation of the reaction;

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a)

If 3 moles of H2 yields 1 mole of NH3

21 moles of H2 will yield 21 × 1 /3 = 7.0 moles of NH3

b)

1 mole of N2 yields 17 g of NH3

3.6 moles of N2 yields 3.6 moles × 17 g of NH3 = 61.2 g of NH3

c)

If 6g of H2 produces 17 g of NH3

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x= 4.15 g of H2

d)

If 6g of hydrogen yields 6.02 × 10^23 molecules of NH3

8.86 × 10^-4g of H2 yields 8.86 × 10^-4g × 6.02 × 10^23 /6 = 8.9 ×10^19 molecules

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