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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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What would happen to the results if the plastic wrap was dirty in a water filtration experiment?​
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3 years ago
8moles of Na2Cr2O2 is how much mass​
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\boxed{\boxed{\mathfrak{ 1\: mole \:of \:Na_2Cr_2O_2\: = \:it's \:Gram\: Mol. \: mass}} }

\underline{ \mathfrak{ Gram \:molecular \:mass \:of \: \red{ Na_2Cr_2O_2}}}

= 2 × 23 + 2 × 52 + 2 × 16

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1 mole of Na_2Cr_2O_2 weighs = 182 g

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=\mathfrak{\blue {\boxed{\underline {1456 \: grams}}}}

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3. The temperature of a fixed mass of gas in a rigid container is decreased from 127.00 degrees
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Answer:

750mmHg

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The following data were obtained from the question:

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T2 = 27°C = 27 +273 = 300K

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400 x P2 = 1000 x 300

Divide both side by 400

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Therefore, the new pressure after cooling is 750mmHg

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2 years ago
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