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spayn [35]
3 years ago
11

Which of the following is the correct name for the compound K2BrO2?

Chemistry
1 answer:
Zigmanuir [339]3 years ago
4 0
The correct name for the molecular formula <span>K2BrO2 would be potassium bromine dioxide. In addition, one of the primary uses of this chemical compound would be in the water purifying, as well as in water treatment industry which could serve as an alternative to chlorine as a disinfectant.</span>
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Na2CO3 + 2HCl ---------&gt; 2NaCl + CO2 + H2O How many moles of NaCl are produced from the reaction of 1.67 x 1022 molecules of
dexar [7]

Answer:

0.0554 moles of NaCl are produced from the reaction of 1.67*10²² molecules of Na₂CO₃ with excess HCl.

Explanation:

The balanced reaction is:

Na₂CO₃ + 2 HCl → 2 NaCl + CO₂ + H₂O

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

  • Na₂CO₃: 1 mole
  • HCl: 2 moles
  • NaCl: 2 moles
  • CO₂: 1 mole
  • H₂O: 1 mole

On the other hand, Avogadro's Number is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023*10²³ particles per mole. Avogadro's number applies to any substance.

In this case, you can apply the following rule of three: if 6.023*10²³ molecules of Na₂CO₃ are contained in 1 mole, 1.67*10²² molecules will be contained in how many moles?

amount of moles=\frac{1.67*10^{22}molecules*1mole }{6.023*10^{23}molecules}

amount of moles= 0.0277 moles

In this case, you can apply the following rule of three: if by stoichiometry 1 mole of Na₂CO₃ produces 2 moles of NaCl, 0.0277 moles of Na₂CO₃ will produce how many moles of NaCl?

amount of moles of NaCl=\frac{0.0277 moles of Na_{2} CO_{3}*2 moles of NaCl }{1 mole of Na_{2} CO_{3}}

amount of moles of NaCl= 0.0554 moles

<u><em>0.0554 moles of NaCl are produced from the reaction of 1.67*10²² molecules of Na₂CO₃ with excess HCl.</em></u>

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