what volume (in ML) of a 0.150 M HNO3 solution is required to completely react wtih 35.7 mL of a 0.108 M Na2CO3 solution according to the following balanced chemical equation?

Na2CO3(aq) + 2 HNO3(aq) --> 2NaNO3(aq) + H2O(l)

then, in the reaction above, what mass (in grams) of carbon dioxide (CO2) is formed?

51.408mL

For your first step it would be 0.108* 0.0357(yours was in mL not L)= 0.0038556

0.0038556molCO2*44g/1mol CO2= 0.170g CO2

thanks for all this help!

so for this one,
1 - it would be 0.108 x 35.7 = 3.86 moles of Na2CO3

2 - 2.86 moles Na2CO3 / 2 Moles HNO3 = 1.93 Moles HNO3.

3 - 1.93 Moles HNO3 x 0.150 M HNO3 = 0.290 L which would equal 290 mL. Is this the final solution?

4 - I need help with this one. I know the molar mass of CO2 is 44. How do i find the moles? (I know this is probably so simple I just can't focus)

1. Calculate moles Na2CO3. moles = M x L.

2. Using the coefficients in the balanced equation, convert moles Na2CO3 to moles HNO3.
3. Now convert moles HNO3 to volume. M = moles/L. YOu have M and moles, solve for L, then convert to mL.

4. For g CO2, step 1 and 2 are the same. Convert mols CO2 to grams. g = mols x molar mass.

Oh, chemistry questions! That's my kind of challenge. Let's break it down, but first, let me grab my lab coat and oversized clown goggles!

To find the volume of the HNO3 solution required, we can use the equation's stoichiometry. From the balanced equation, we see that one mole of Na2CO3 reacts with two moles of HNO3.

Given the volume of the Na2CO3 solution (35.7 mL) and its concentration (0.108 M), we can determine the number of moles of Na2CO3 using the formula:

moles of Na2CO3 = concentration * volume
moles of Na2CO3 = 0.108 M * 0.0357 L
moles of Na2CO3 = 0.0038616 mol

Since the stoichiometry ratio tells us that two moles of HNO3 react with one mole of Na2CO3, we know that twice the number of moles of Na2CO3 will be required for HNO3.

moles of HNO3 = 2 * moles of Na2CO3
moles of HNO3 = 2 * 0.0038616 mol
moles of HNO3 = 0.0077232 mol

Finally, we can calculate the volume of the HNO3 solution using its molarity:

volume of HNO3 = moles of HNO3 / concentration
volume of HNO3 = 0.0077232 mol / 0.150 M
volume of HNO3 = 0.05149 L
volume of HNO3 = 51.49 mL

So, approximately 51.49 mL of the HNO3 solution is required to completely react with 35.7 mL of the Na2CO3 solution.

Now, let's dive into the second part of your question, the mass of CO2 formed. From the balanced equation, we can see that the stoichiometric ratio between Na2CO3 and CO2 is 1:1.

Given that we've used 0.0038616 mol of Na2CO3, we can determine the number of moles of CO2 formed:

moles of CO2 = moles of Na2CO3
moles of CO2 = 0.0038616 mol

To find the mass of CO2, we need to multiply the number of moles by its molar mass, which is approximately 44.01 g/mol:

mass of CO2 = moles of CO2 * molar mass of CO2
mass of CO2 = 0.0038616 mol * 44.01 g/mol
mass of CO2 ≈ 0.1701 g

Therefore, approximately 0.1701 grams of carbon dioxide (CO2) is formed in the reaction.

Just remember, chemistry can be a balancing act, just like a clown riding a unicycle while juggling test tubes!

To find the volume of the HNO3 solution required to completely react with the Na2CO3 solution, we can use the concept of stoichiometry. Stoichiometry is the calculation of the quantities of reactants and products involved in a chemical reaction.

1. Start by writing out the balanced chemical equation for the reaction:
Na2CO3(aq) + 2 HNO3(aq) --> 2 NaNO3(aq) + H2O(l)

2. Determine the stoichiometric ratio between Na2CO3 and HNO3 from the balanced equation. In this case, the ratio is 1:2. This means that for every 1 mole of Na2CO3, we need 2 moles of HNO3.

3. Convert the given volume of the Na2CO3 solution into moles. Use the formula:
Moles = Concentration (M) x Volume (L)
For the Na2CO3 solution, Moles = 0.108 M x 0.0357 L = 0.0038616 mol

4. Use the stoichiometric ratio to determine the moles of HNO3 required. Since the ratio is 1:2, twice the moles of Na2CO3 is needed. So, 2 x 0.0038616 mol = 0.0077232 mol of HNO3.

5. Find the volume of the HNO3 solution using its concentration and the moles calculated in the previous step. Rearrange the formula from step 3 to solve for volume:
Volume (L) = Moles / Concentration (M)
Volume = 0.0077232 mol / 0.150 M ≈ 0.05149 L

6. Convert the volume from liters to milliliters:
Volume (mL) = 0.05149 L x 1000 mL/L ≈ 51.49 mL

Therefore, approximately 51.49 mL of the 0.150 M HNO3 solution is required to completely react with 35.7 mL of the 0.108 M Na2CO3 solution.

Now, let's find the mass of carbon dioxide (CO2) formed.

7. Use the moles of Na2CO3 calculated in step 3 to determine the moles of CO2 formed. From the balanced equation, the stoichiometric ratio between Na2CO3 and CO2 is 1:1. Therefore, the moles of CO2 formed will also be 0.0038616 mol.

8. Calculate the molar mass of CO2, which is the sum of the atomic masses of carbon (C) and two oxygens (O):
Molar mass (CO2) = (12.01 g/mol) + 2 × (16.00 g/mol) ≈ 44.01 g/mol

9. Use the moles and molar mass of CO2 to calculate the mass using the formula:
Mass (g) = Moles × Molar mass
Mass = 0.0038616 mol × 44.01 g/mol ≈ 0.170 g

Therefore, approximately 0.170 grams of carbon dioxide (CO2) is formed in the given reaction.