Within this experiment we are calculating the limiting reactant from mol of Ch3C

ID: 898905 • Letter: W

Question

Within this experiment we are calculating the limiting reactant from mol of Ch3CooH and mol of NaHCO3.

How I found the limiting reactant.

A) Mols of CH3COOH =((5mlCH3COOHx8.3gCH3COOH/10ml)/0.05(%))/60.052 g/mol CH3COOH =.0034 mol

then I took

B) Mols NaHCO3 = (0.1g NaHCO3 X1mol/84.007g NaHCO3) =.0011 mol

I took these values and subtracted A-B. .0034 mol - .0011 mol

So the limiting reactant would be NaHCO3 correct? ( think this is how you do it but I could be wrong)

From here I need to find the theoretical yield of CO2 (mol) and in (g).

Please help and state how you found these values, I have to do this for 5 other problems so i want to make sure I am doing this correctly.

Let me know if I need to add more information! I will give good ratings!

Thank you very much!

NaHCO3 + CH3COOH = NaCH3COO + H2O + CO2

1 mole of NaHCO3 reacts with 1 mole of CH3COOH

According to ypur calculations,

There are 0.0034 moles of CH3COOH and 0.0011 moles of NaHCO3.

0.0011 moles of NaHCO3 reacts with 0.0011 moles of CH3COOH. After that no NaHCO3 is left.

Hence NaHCO3 is the limiting reagent

Excess NaHCO3 = 0.0034 - 0.0011 = 0.0023 moles

1 mole of NaHCO3 produces 1 mole of CO2.

Hence 0.0011 moles of NaHCO3 produces 0.0011 moles of CO2

Mass of CO2 = 0.0011 x 44 = 0.0484 g

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