Question

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?

Answer 1

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.

Answer 2

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)

Answer 3

51.408mL

Answer 4

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