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joja [24]
2 years ago
5

How many grams of carbon (C) would be present in carbon monoxide (CO) that contains 2.666 grams of oxygen (O)? For example if th

e sample size was 10 L or 10,000 L.
Chemistry
1 answer:
oee [108]2 years ago
5 0

Given :

Mass of oxygen containing carbon monoxide (CO) is 2.666 gram .

To Find :

How many grams of carbon (C) would be present in carbon monoxide (CO) that contains 2.666 grams of oxygen (O) .

Solution :

By law of constant composition , a given chemical compound always contains its component elements in fixed ratio (by mass) and does not depend on its source and method of preparation.

So , volume of solution does not matter .

Moles of oxygen , n=\dfrac{2.666}{16}=0.167\ mole .

Now , molecule of CO contains 1 mole of C .

So , moles of C is also 0.167 mole .

Mass of carbon , m=12\times 0.167=2\ g .

Therefore , mass of carbon is 2 grams .

Hence , this is the required solution .

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A solution contains an unknown amount of dissolved magnesium. Addition of
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Taking into account the reaction stoichiometry, 2.13 grams of magnesium was dissolved in the solution.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

Mg²⁺(aq) + Na₂CO₃(aq) → MgCO₃(s) + 2 Na⁺(aq)

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • Mg²⁺: 1 mole
  • Na₂CO₃: 1  mole
  • MgCO₃: 1 mole
  • Na⁺: 2 moles

The molar mass of the compounds is:

  • Mg²⁺: 24.3 g/mole
  • Na₂CO₃: 106 g/mole
  • MgCO₃: 84.3 g/mole
  • Na⁺: 23 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • Mg²⁺: 1 mole ×24.3 g/mole= 24.3 grams
  • Na₂CO₃: 1 mole ×106 g/mole= 106 grams
  • MgCO₃: 1 mole ×84.3 g/mole=84.3 grams
  • Na⁺: 2 moles ×23 g/mole= 46 grams

<h3>Mass of magnesium dissolved</h3>

The following rule of three can be applied: If by reaction stoichiometry 1 mole of Na₂CO₃ react with 24.3 grams of magnesium, 0.0877 moles of Na₂CO₃ react with how much mass of magnesium?

mass of magnesium=\frac{0.0877 moles of Na_{2}C O_{3}x24.3 grams of magnesium }{1 mole of Na_{2}C O_{3}}

<u><em>mass of magnesium= 2.13 grams</em></u>

Finally, 2.13 grams of magnesium was dissolved in the solution.

Learn more about the reaction stoichiometry:

<u>brainly.com/question/24741074</u>

<u>brainly.com/question/24653699</u>

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