Chemistry Help?

Question

1) What volume of 3 M HCl would be required to react with 5 g of a mixture that is 50% by mass sodium carbonate and 50 % sodium hydrogen carbonate?

2) The NaCl was not dried suffieciently- would the percent yield be higher or lower than the theoretical yield?

3) If small portions of NaCl were lost to spattering during the evaporation of the solvent in this experiment, would the percentage yield be higher or lower? Explain.

Answer 1

[HCl] = 3 M
m(Na2CO3) = 5 g
Find the mole for each of them first:
n = m/M
= 5/106
= 0.0472 mol

n(HCl) = n(Na2CO3) as according to the equation 2 mole of HCl reacts with 1 mole of Na2CO3 so the molar ratio is 2:1

Find the Volume required

using the formula n = CV 

Here n = 2(0.0472) and C = 3 

Therefore V = (0.0472/3)2 = 0.0315 L
= 31.5 mL