What volume (in mL) of a 0.125 MHNO3 solution is required to completely react with 26.1 mL of a 0.112 MNa2CO3 solution according to the following balanced chemical equation?
Na2CO3(aq)+2HNO3(aq)→2NaNO3(aq)+CO2(g)+H2O(l)
Na2CO3(aq)+2HNO3(aq)→2NaNO3(aq)+CO2(g)+H2O(l)
How many mols of Na2CO3 do you have? That's mols = M x L = 26.1 mL of a 0.112 MNa2CO3 soln or 0.112 M x 0.0261 L = ?
Now convert to mols HNO3. mols HNO3 will be 2 x mols Na2CO3, then
M = mols/L to convert to volume of HNO3 needed. M = mols/L. You know mols HNO3, you know M HNO3, solve for L HNO3 and convert to mL if desired. Post your work if you get stuck.
To find the volume of the HNO3 solution required, we need to use the concept of stoichiometry and the balanced chemical equation.
The balanced chemical equation states that for every 1 mole of Na2CO3, 2 moles of HNO3 are required to completely react. This means that the mole ratio between Na2CO3 and HNO3 is 1:2.
First, let's convert the mL of Na2CO3 solution to moles using its concentration:
Moles of Na2CO3 = Volume (in L) × Concentration (in M)
= (26.1 mL ÷ 1000) L × 0.112 M
= 0.0029244 moles of Na2CO3
Since the mole ratio between Na2CO3 and HNO3 is 1:2, we can determine the moles of HNO3 required:
Moles of HNO3 = 2 × Moles of Na2CO3
= 2 × 0.0029244 moles
= 0.0058488 moles of HNO3
Now, we can find the volume of the HNO3 solution required by dividing the moles of HNO3 by its concentration:
Volume (in L) = Moles of HNO3 ÷ Concentration (in M)
= 0.0058488 moles ÷ 0.125 M
= 0.0467912 L
Finally, we convert the volume from liters to milliliters:
Volume (in mL) = Volume (in L) × 1000
= 0.0467912 L × 1000
= 46.7912 mL
Therefore, approximately 46.8 mL of the 0.125 M HNO3 solution is required to completely react with 26.1 mL of the 0.112 M Na2CO3 solution.