Calculate the number of moles of 2NaCl, that can be obtained from 25g of Na2CO3 in the presence of excess HCl

Answer 1

To solve this problem, we need to calculate the number of moles of Na2CO3 first, and then use the stoichiometry of the balanced chemical equation to determine the number of moles of 2NaCl that can be obtained.

1. Calculate the molar mass of Na2CO3:
Na: 2 * 22.99 g/mol = 45.98 g/mol
C: 1 * 12.01 g/mol = 12.01 g/mol
O: 3 * 16.00 g/mol = 48.00 g/mol

Total molar mass of Na2CO3 = 45.98 + 12.01 + 48.00 = 105.99 g/mol

2. Calculate the number of moles of Na2CO3:
Moles of Na2CO3 = Mass of Na2CO3 / Molar mass of Na2CO3
Moles of Na2CO3 = 25 g / 105.99 g/mol = 0.236 moles

3. Use the stoichiometry of the balanced chemical equation to determine the number of moles of 2NaCl:
The balanced chemical equation for the reaction is:
Na2CO3 + 2HCl -> 2NaCl + H2O + CO2

From the balanced equation, we can see that 1 mole of Na2CO3 produces 2 moles of 2NaCl.

Therefore, the number of moles of 2NaCl = 2 * Moles of Na2CO3
Number of moles of 2NaCl = 2 * 0.236 moles = 0.472 moles

So, 25g of Na2CO3 can produce 0.472 moles of 2NaCl.