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Aliun [14]
3 years ago
8

Determine the AMOUNT OF NO2, LIMITING REACTANT, AND THE AMOUNT AND NAME OF EXCESS REACTANT.

Chemistry
1 answer:
zepelin [54]3 years ago
3 0

Answer:

balanced equation mole ratio 5 2 mol NO/1 mol O2

10.00 g O2 3 1 mol O2/32.00 g O2 5 0.3125 mol O2

20.00 g NO 3 1 mol NO/30.01 g NO 5 0.6664 mol NO

actual mole ratio 5 0.6664 mol NO/0.3125 mol O2 5 2.132 mol NO/1.000 mol O2

Because the actual mole ratio of NO:O2 is larger than the balanced equation mole

ratio of NO:O2, there is an excess of NO; O2 is the limiting reactant.

Mass of NO used 5 0.3125 mol O2 3 2 mol NO/1 mol O2 5 0.6250 mol NO

0.6250 mol NO 3 30.01 g NO/1 mol NO 5 18.76 g NO

Mass of NO2 produced 5 0.6250 mol NO2 3 46.01 g NO2/1 mol NO2 5 28.76 g NO2

Excess NO 5 20.00 g NO 2 18.76 g NO 5 1.24 g N

Explanation:

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If a jet's cruising altitude is 32,200 ft (to three significant figures), this distance in km is: (1 mile = 1.61 km; 1 mile = 52
DochEvi [55]

If a jet's cruising altitude is 32,200 ft (to three significant figures), this distance in km is 9.82 km.

<h3>What are significant figures?</h3>

Significant figures are the number of digits in a value, often a measurement, that contributes to the degree of accuracy of the value. We start counting significant figures at the first non-zero digit.

In this case, given the conversion factors from miles to kilometres and from miles to feet, we can directly compute the jet’s cruising altitude in kilometres as shown below:

32200 ft X \frac{1 mile}{5280 ft}  X \frac{1.61km}{1 mile}

=9.82 km

Learn more about the significant figures here:

brainly.com/question/14359464

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