Question

What volume (in mL) of a 0.150 M HNO3 solution will completely react with 35.7 mL of a 0.108 M Na2CO3 solution according to the following balanced chemical equation?

Na2CO3(aq)+2HNO3(aq)→2NaNO3(aq)+CO2(g)+H2O(l)

In the reaction in Part A, what mass (in grams) of carbon dioxide forms?

Answer 1

51.43 mL of 0.150 M HNO3 solution is required to react with 35.7 mL of 0.108 M Na2CO3 solution and will produce 0.1697 grams of carbon dioxide (CO2).

Step-by-step explanation:

To determine the volume of 0.150 M HNO3 solution needed to react with 35.7 mL of a 0.108 M Na2CO3 solution, we use the stoichiometry of the balanced chemical equation. The equation states that one mole of Na2CO3 reacts with two moles of HNO3.

First, we calculate the moles of Na2CO3:

• Moles of Na2CO3 = 0.108 M × 0.0357 L = 0.003857 moles Na2CO3

From the balanced equation, 1 mole of Na2CO3 requires 2 moles of HNO3, so:

• Moles of HNO3 required = 2 × moles of Na2CO3
• Moles of HNO3 required = 2 × 0.003857 moles = 0.007714 moles HNO3

Now, to find the volume of HNO3 solution needed (using its concentration):

• Volume of HNO3 = moles of HNO3 / concentration of HNO3
• Volume of HNO3 = 0.007714 moles / 0.150 M = 0.05143 L
• Volume of HNO3 in mL = 0.05143 L × 1000 mL/L = 51.43 mL

Therefore, 51.43 mL of 0.150 M HNO3 solution is required to completely react with 35.7 mL of 0.108 M Na2CO3 solution.

To find the mass of carbon dioxide (CO2) that forms from the reaction of 35.7 mL of a 0.108 M Na2CO3 solution, we first find the moles of Na2CO3 (as calculated before, 0.003857 moles). Since the reaction produces 1 mole of CO2 for every mole of Na2CO3 reacted, the moles of CO2 formed will also be 0.003857.

Next, we calculate the mass of CO2 using the molar mass of CO2 (44.01 g/mol):

• Mass of CO2 = moles of CO2 × molar mass of CO2
• Mass of CO2 = 0.003857 moles × 44.01 g/mol = 0.1697 grams

The mass of CO2 formed in this reaction is 0.1697 grams.

Answer 2

First part: volume of HNO3 solution = 51.4 mL

Second part: mass of CO2 = 0.170 g

Step-by-step explanation:

First part:

Based on the volume and concentration of the Na2CO3 solution, there is 3.86 mmol of Na2CO3 available to react with HNO3 . Considering the stoichiometric relationship, you then calculate the volume of a 0.150 M solution that contains 7.71 mmol of HNO3 .

51.4 mL

Second part:

Since the reactant and product have a stoichiometric relationship that is 1:1, 0.00386 mol of Na2CO3 can react to form 0.00386 mol of CO2 . The molar mass of CO2 (44.01 g/mol ) is then used to convert the moles to the corresponding mass in grams.

0.170 g