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galben [10]
1 year ago
15

In an experiment magnesium ribbon was heated in air. The product was found to be heavier than the original ribbon. Potassium man

ganate 7 was on the other hand, heated in air and product formed was to be lighter. Explain the difference on the observation made
Chemistry
1 answer:
KatRina [158]1 year ago
7 0

Burning a magnesium ribbon in the air is an addition reaction while heating potassium manganate 7 is a decomposition reaction.

<h3>Addition and decomposition reactions</h3>

Magnesium burns in air to produce magnesium oxide as follows:

2Mg + O_2 --- > 2MgO

Potassium manganate 7 burns to produce multiple products as follows:

2 KMnO_4 --- > K_2MnO_4 + MnO_2(s) + O_2

Thus, the MgO will be heavier than Mg. On the other hand, MnO_2 will be less heavy than KMnO_4.

More on reactions can be found here: brainly.com/question/17434463

#SPJ1

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3 years ago
If you mix a 25.0mL sample of a 1.20m potassium sulfide solution with 15.0 mL of a 0.900m basium nitrate solution and the reacti
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Answer:

Limiting reagent: barium nitrate

Theoretical yield: 2.29 g BaS

Percent yield: 87%

Explanation:

The corrected balanced reaction equation is:

K₂S + Ba(NO₃)₂ ⇒ 2KNO₃ + BaS

The amount of potassium sulfide added is:

(25.0 mL)(1.20mol/L) = 30 mmol

The amount of barium nitrate added is:

(15.0mL)(0.900mol/L = 13.5 mmol

Since the reactant react in a 1:1 molar ratio, barium nitrate is the limiting reagent. If all of the limiting reagent reacts, the amount of barium sulfide produced is:

(13.5 mmol Ba(NO₃)₂)(BaS/Ba(NO₃)₂ ) = 13.5 mmol BaS

Converting this amount to grams gives the theoretical yield of BaS (molar mass 169.39 g/mol).

(13.5 mmol)(169.39 g/mol)(1g/1000mg) = 2.29 g BaS

The percent yield is calculated as follows:

(actual yield) / (theoretical yield) x 100%

(2.0 g) / (2.29 g) x 100% = 87%

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