Help chemistry

Hi Stephanie hammond

The answer is 94.3 mL.

First, we need to calculate the number of moles of sodium carbonate using its molar mass:

molar mass of Na2CO3 = 2(22.99 g/mol) + 12.01 g/mol + 3(16.00 g/mol) = 105.99 g/mol

moles of Na2CO3 = mass / molar mass = 0.500 g / 105.99 g/mol = 0.004713 mol

According to the balanced chemical equation, 1 mole of Na2CO3 reacts with 2 moles of HCl. Therefore, we need twice as many moles of HCl as Na2CO3 to completely react:

moles of HCl = 2 × moles of Na2CO3 = 2 × 0.004713 mol = 0.009426 mol

Now we can use the molarity of hydrochloric acid to calculate the required volume:

Molarity = moles / volume

volume = moles / Molarity = 0.009426 mol / 0.100 mol/L = 0.09426 L

Therefore, the volume of 0.100 M hydrochloric acid required to react completely with 0.500 g of sodium carbonate is 0.09426 L, or 94.26 mL.