How many mL of 0.1 M HCl are required to react completely with 1 g mixture Of Na2CO3 and NaHCO3 containing equimolar amounts of both?Read more on Sarthaks.com - https://www.sarthaks.com/630419/many-required-react-completely-mixture-na2co3-nahco3-containing-equimolar-amounts-both

First, we need to understand the reactions that are taking place.

1. Na2CO3 + 2HCl -> 2NaCl + H2O + CO2
2. NaHCO3 + HCl -> NaCl + H2O + CO2

From these reactions, we can see that one mole of Na2CO3 reacts with 2 moles of HCl and one mole of NaHCO3 reacts with 1 mole of HCl.

Next, we need to calculate the number of moles in 1g of the mixture. Since the mixture contains equimolar amounts of Na2CO3 and NaHCO3, we can say that 0.5g is Na2CO3 and 0.5g is NaHCO3.

The molar mass of Na2CO3 is approximately 105.99 g/mol and the molar mass of NaHCO3 is approximately 84.01 g/mol.

So, the number of moles of Na2CO3 = 0.5g / 105.99 g/mol = 0.0047 moles And, the number of moles of NaHCO3 = 0.5g / 84.01 g/mol = 0.006 moles

From the reactions, we know that 1 mole of Na2CO3 requires 2 moles of HCl and 1 mole of NaHCO3 requires 1 mole of HCl. Therefore, the total moles of HCl required = 20.0047 + 10.006 = 0.0154 moles

Since the molarity of the HCl solution is 0.1M, which means there are 0.1 moles of HCl in 1L of solution, the volume of HCl solution required can be calculated by:

Volume = Moles / Molarity = 0.0154 moles / 0.1 M = 0.154 L = 154 mL

So, 154 mL of 0.1 M HCl are required to react completely with 1 g mixture of Na2CO3 and NaHCO3 containing equimolar amounts of both.