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Tpy6a [65]
3 years ago
6

How many grams of ammonia must you start with to make 900.00 l of a 0.140 m aqueous solution of nitric acid? assume all the reac

tions give 100% yield?
Chemistry
1 answer:
bogdanovich [222]3 years ago
3 0
You need the set of reactions that goes from ammonia to nitric acid.
<span>
1) 4NH3(g)+5O2(g)-->4NO(g)+6H2O(g)

2) 2NO(g)+O2(g)-->2NO2(g)

3) 3NO2(g)+H2O(l)-->2HNO3(aq)+NO(g)

State the ratio of moles of HNO3 to NH3:

4 moles of NH3 produce 4 mole of NO,

4 moles of NO produce 4 moles of NO2

4 moles of NO2 produce 4 * (2 / 3) moles of HNO3 = 8/3 moles of HNO3.

=> (8/3) moles HNO3 : 4 moles NH3

Calculate the number of moles of HNO3 in 900.00 l of 0.140 M solution

M = n / V => n = M * V = 0.140 M * 900.00 liter = 126 moles HNO3

Use proportions:

(</span><span>8/3) moles HNO3 / 4 moles NH3 = 126 moles HNO3 / x

=> x = 126 moles HNO3 * 4 moles NH3 / (8/3 moles HNO3) = 189 moles NH3

Convert moles to grams:

molar mass NH3 = 14 g/mol + 3 * 1g/mol = 17 g/mol

mass in grams = number of moles * molar mass = 189 moles * 17 g/mol = 3213 g

Answer: 3213 g.
</span>
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Answer:

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Explanation:

Full question:

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3 years ago
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Answer: grams=0.048g, ounces=0.0017oz, 0.00011lb

Explanation:

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48 mg x  1 g

          ÷  1000 mg  = 0.048 g

48 mg x  1 g            x 16 oz        

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48 mg x  1 g            x 1 lb        

          ÷ 1000 mg  ÷ 453.6 g     = 0.00011 lb

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