What volume (in mL) of a 0.300 M HNO3 solution will completely react with 43.9 mL of a 0.146 M Na2CO3 solution according to this balanced chemical equation?

Na2CO3(aq)+2 HNO3(aq)->2 NaNO3(aq)+CO2(g)+H20(l)

mols Na2CO3 = M x L = ?

Using the coefficients in the balanced equation, convert mols Na2CO3 to mols HNO3.
Now M HNO3 = mols HNO3/L HNO3 solution.
You know M and mols, solve for L and convert to mL.

To determine the volume of the HNO3 solution needed to react with the given volume of Na2CO3 solution, we can use the mole-to-mole ratio derived from the balanced chemical equation.

1. Calculate the moles of Na2CO3 solution:
moles of Na2CO3 = volume (L) × concentration (M)
moles of Na2CO3 = 0.0439 L × 0.146 M

2. Use the mole ratio between Na2CO3 and HNO3 to determine the moles of HNO3 required. From the balanced equation, the mole ratio between Na2CO3 and HNO3 is 1:2.
moles of HNO3 = 2 × moles of Na2CO3

3. Calculate the volume of HNO3 solution:
volume (L) = moles / concentration
volume (L) = (2 × moles of Na2CO3) / 0.300 M

4. Convert the volume from liters to milliliters:
volume (mL) = volume (L) × 1000

Let's plug in the values and calculate:

moles of Na2CO3 = 0.0439 L × 0.146 M = 0.0064054 mol

moles of HNO3 = 2 × 0.0064054 mol = 0.0128108 mol

volume (L) = (0.0128108 mol) / 0.300 M = 0.0427033 L

volume (mL) = 0.0427033 L × 1000 = 42.7033 mL

Therefore, approximately 42.7 mL of the 0.300 M HNO3 solution will be needed to completely react with 43.9 mL of the 0.146 M Na2CO3 solution.