Question

(i)What mass of carbon dioxide (in kg) is produced upon the complete combustion of

18.9 L of propane (approximate contents of one 5-gallon tank)? Assume that the


density of the liquid propane in the tank is 0.621 g/mL . (Hint: Begin by writing

a balanced equation for the combustion reaction.)

m=? kg



(ii) A 22.0 - mL sample of an unknown H3PO4 solution is titrated with a 0.110 M NaOH

solution. The equivalence point is reached when 18.60 mL of NaOH solution is added.


What is the concentration of the unknown H3PO4 solution? The neutralization reaction is as

follows:

H3PO4(aq) + 3NaOH(aq) ---> 3H2O(l) + Na3PO4(aq)


M=?

Answer 1

The answer to your question is:

a) 35211 g of CO2

b) Molarity = 0.03

Step-by-step explanation:

a)

mass of CO2 = ?

Volume = 18.9 L

density = 0.621 g/ml

Reaction

C₃H₈ + 5O₂ ⇒ 3CO₂ + 4H₂O

Density = mass / volume

mass = density x volume

mass = 0.621 x 18900

mass = 11737 g

MW of C3H8 = 44 g

MW of 3CO2 = 132 g

44 g of C3H8 ------------------- 132 g of CO2

11737 g of C3H8 -------------------- x

x = (11737 x 132) / 44

x = 35211 g of CO2

b) H3PO4(aq) + 3NaOH(aq) ---> 3H2O(l) + Na3PO4(aq)

22 ml 18.6 ml

[?] 0.11 M

Calculate moles of NaOH

n = M x V

n = 0.11 x 0.0186

n = 0.0020 moles

From the balance reaction

1 mol of H3PO4 ------------------ 3 moles of NaOH

x ------------------ 0.0020 moles

x = (0.002 x 1) / 3

x = 0.0007 mol of H3PO4

Molarity = 0.0007 / 0.022

Molarity = 0.03