8 What volume of a 0125M HCI will completely react with 155g

8. What volume of a 0.125M HCI will completely react with 15.5g of Na2CO3? The balanced reaction is: 2 HCI(aj + Na2CO3(s) H2Og + CO2(g) + 2 NaClaq)

Number of moles of Na2CO3 = 15.5 g / 105.988 g/mol = 0.146 mol

from the balanced equation we can say that

1 mole of Na2CO3 requires 2 mole of HCl so

0.146 mole of Na2CO3 will require

= 0.146 mole of Na2CO3 *(2 mole of HCl / 1 mole of Na2CO3)

= 0.292 mole of HCl

molarity = number of moles of HCl / volume of solution in L

volume of solution in L = 0.292 mol / 0.125 M = 2.34 L

1 L = 1000 mL

2.34 L = 2340 mL

Therefore, volume of solution would be 2340 mL