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)

To find the volume of the HNO3 solution, we need to use the stoichiometric ratio from the balanced chemical equation.

From the equation, we can see that 2 moles of HNO3 react with 1 mole of Na2CO3. We are given the concentration (Molarity) of the Na2CO3 solution, so we need to calculate the number of moles of Na2CO3 in the given volume.

Number of moles of Na2CO3 = concentration × volume
= 0.146 M × 43.9 mL
= 6.4034 × 10^(-3) mol

Now, using the stoichiometric ratio, we can calculate the number of moles of HNO3 required.

Number of moles of HNO3 = 2 × Number of moles of Na2CO3
= 2 × 6.4034 × 10^(-3) mol
= 1.2807 × 10^(-2) mol

Finally, we can calculate the volume of the HNO3 solution using the concentration and number of moles of HNO3.

Volume of HNO3 solution = Number of moles of HNO3 / concentration
= 1.2807 × 10^(-2) mol / 0.300 M
≈ 0.0427 L
= 42.7 mL

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

To solve this problem, we need to use the concept of stoichiometry, which involves the balanced chemical equation, molarity, and volume relationships.

Given:
Molarity of HNO3 solution (concentration) = 0.300 M
Volume of HNO3 solution = unknown (let's say V mL)
Molarity of Na2CO3 solution (concentration) = 0.146 M
Volume of Na2CO3 solution = 43.9 mL

From the balanced chemical equation:
1 mole of Na2CO3 reacts with 2 moles of HNO3

Step 1: Find the number of moles of Na2CO3.
Using the volume (V) and concentration (C) formula: moles = volume (L) × concentration (M). It's important to convert mL to L for volume consistency.
moles of Na2CO3 = (43.9 mL ÷ 1000 mL/L) × 0.146 M

Step 2: Convert moles of Na2CO3 to moles of HNO3.
Since the balanced chemical equation shows that one mole of Na2CO3 reacts with 2 moles of HNO3, we can write the conversion factor: (2 moles HNO3 / 1 mole Na2CO3)
moles of HNO3 = moles of Na2CO3 × (2 moles HNO3 / 1 mole Na2CO3)

Step 3: Convert moles of HNO3 to volume (in mL) of HNO3.
Using the volume (V) and concentration (C) formula: volume = moles ÷ concentration.
volume (V) of HNO3 = (moles of HNO3 ÷ 0.300 M) × 1000 mL/L

Calculating each step:
- Step 1: moles of Na2CO3 = (43.9 mL ÷ 1000 mL/L) × 0.146 M
- Step 2: moles of HNO3 = moles of Na2CO3 × (2 moles HNO3 / 1 mole Na2CO3)
- Step 3: volume (V) of HNO3 = (moles of HNO3 ÷ 0.300 M) × 1000 mL/L

Let's calculate the answer:
Step 1: moles of Na2CO3 = (43.9 mL ÷ 1000 mL/L) × 0.146 M
= 0.00640384 moles

Step 2: moles of HNO3 = 0.00640384 moles × (2 moles HNO3 / 1 mole Na2CO3)
= 0.01280768 moles

Step 3: volume (V) of HNO3 = (0.01280768 moles ÷ 0.300 M) × 1000 mL/L
= 42.692 mL

Therefore, the volume of a 0.300 M HNO3 solution required to react completely with 43.9 mL of a 0.146 M Na2CO3 solution is approximately 42.692 mL.