calculate the moles of sulfuric acid that remained in the solution after reaction with the carbonate.

the tared mass of carbonate is 1.5179g

You didn't post all of the problem.

i've calculated the moles of sulfuric acid present in the 50ml of standard acid solution to be 2.378x10^-1mol. the average volume of sodium hydroxide required for the back titration to be 31.03mL. and the moles of sodium hydroxide required for the back titration to be 2.951x10^-3. I'm not sure how to proceed

You've still posted portions although the last helps. My guess is that you added an excess of a standard H2SO4 soln to a CO3^= unknown and you're back titrating to find amount of excess H2SO4 added.

total mols H2SO4 added initially minus 1/2 the mols NaOH to back titrate gives you mols carbonate ion there originally in the unknown.

im guessing im missing the concentration of H2SO4 is 0.04755M and concentration for sulfuric acid is 0.09510M. In the flash i have 50.00ml of sulfuric acid and 25.00ml of impure sodium carbonate solution.

To calculate the moles of sulfuric acid remaining in the solution after the reaction with the carbonate, you will need to know the reactant stoichiometry and the balanced equation for the reaction.

1. Write the balanced equation for the reaction between sulfuric acid (H2SO4) and the carbonate:

H2SO4 + CO3 → H2O + CO2 + SO2

2. Determine the molar mass of sulfuric acid (H2SO4):

H2SO4:
(2 x 1.008 g/mol (H)) + 32.06 g/mol (S) + (4 x 16.00 g/mol (O)) = 98.09 g/mol

3. Convert the mass of the carbonate to moles:

Given tared mass of carbonate = 1.5179 g

Molar mass of carbonate:
(Molar mass of C) + (3 x Molar mass of O)
= 12.01 g/mol + (3 x 16.00 g/mol)
= 60.01 g/mol

Moles of carbonate = (Given mass of carbonate) / (Molar mass of carbonate)
= 1.5179 g / 60.01 g/mol

4. Determine the stoichiometric ratio between sulfuric acid and carbonate:

From the balanced equation in step 1, the stoichiometric ratio between sulfuric acid and carbonate is 1:1.

5. Calculate the moles of sulfuric acid that reacted:

Moles of sulfuric acid reacted = Moles of carbonate (from step 3)

6. Calculate the moles of sulfuric acid remaining in the solution:

Since the stoichiometric ratio between sulfuric acid and carbonate is 1:1, the moles of sulfuric acid remaining in the solution will also be equal to the moles of sulfuric acid that reacted (from step 5).

The moles of sulfuric acid that remained in the solution after reaction with the carbonate can be calculated using the given data and the steps outlined above.