Question

Cane CHM dy Ch Sng 201 S Postlab Questions [show your work in the space provided] Nazcozei05.99 Consider the following equation: 1. 2H,PO,+3Na,CO, 3CO,+3H,O+ 2Na,PO, a. How many moles of CO, are expected from 10.0 g of Na,CO,7 b. How many moles of CO, are expected from 100.0 mL of 0.20 M H,PO c. What is the limiting reagent if a. and b. are reacted? d. How many moles of CO, are expected from the reaction? What volume of CO, is expected from the reaction? Assume that e. pressure 1 atm and the temperature is 25 degrees C. CO cO CHCOH 2. List two reasons that explain why the volume of CO2 produced in our reactions might not be the same as the predicted volume of CO2 a) At what flask (1-6) did the diameter of the balloons start to stay about 3. the same? 9Tom b) At what flask (1-6) did the sodium carbonate completely dissolve? 19 c) What is the significance of both of these observations? Student Name Date

Answer 1

Given reaction -

1) 3 H₃PO₄ + 3 Na₂CO₃ ----> 3 CO₂ + 3 H₂O + 2 Na₃PO₄

a) First we find the molecular wt. of Na₃PO₄

Molecular wt. of Na = 22.99

Molecular wt. of C = 12.01

Molecular wt. of O = 16

Molecular wt. of Na₃PO₄ = 3*22.99+12.01+3*16 = 105.99

Moles of Na₃PO₄ in 10g of Na₃PO₄ = 105.99/10 = 0.0943 moles

As per given reaction, 3 moles of Na₃PO₄ give 3 moles of CO₂. Hence their stoichiometric ratio is 1:1

Therefore moles of CO₂ produced from 10g of Na₃PO₄ is 0.0943 moles

b) Molarity = no. of moles / volume in (L)

So, 0.2 M H₃PO₄ = moles of H₃PO₄ / volume of H₃PO₄

0.2 ₌ moles ofH₃PO₄ / (100 * 10^-3)

So moles ofH₃PO₄ = 0.02 moles

As per given reaction, 2 moles of H₃PO₄ give 3 moles of CO₂. Hence their stoichiometric ratio is 2:3

So - 0.02 moles of H₃PO₄ * 3 moles of CO₂ / 2 moles of H₃PO₄

= 0.03 moles of CO₂

c) As per (question a) , there are 0.0943 moles of  Na₃PO₄

As per (question b) there are 2 moles of H₃PO₄

To find limiting reagent:

From stoichiometry we know that moles of Na₃PO₄  to moles of CO₂ ratio is 3:3

So - 0.0943 moles of Na₃PO₄ * 3 moles of CO₂ / 3 moles of Na₃PO₄

= 0.0943 moles of CO₂

From stoichiometry we know that moles of H₃PO₄  to moles of CO₂ ratio is 2:3

So - 0.02 moles of H₃PO₄ * 3 moles of CO₂ / 2 moles of H₃PO₄

= 0.03 moles of CO₂

Therefore since H₃PO₄ limits the production of CO₂, it is the limiting reactant.

d) As seen from above , 0.03 moles of CO₂ are expeced from the reaction.

e) P= 1 atm

T = 25 deg C = 25+273.15 = 298.15 K

R= 0.08205 L atm/ mol K

As per ideal gas equation - -> PV = nRT

Therefore V = nRT/P

= 0.03*0.08205*298.15 / 1

V = 0.7339 L