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Na2CO3 + 2HCl ---------> 2NaCl + CO2 + H2O How many moles of NaCl are produced from the reaction of 1.67 x 1022 molecules of Na2CO3 with excess HCl?

User Paras Gupta
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18 votes

Answer:

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

Step-by-step 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=(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=(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

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

User Jacob Ras
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