Question

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?

Answer 1

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

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