(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=?

The answer to your question is:

a) 35211 g of CO2

b) Molarity = 0.03

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

answer: at room temperature (roughly 20 degrees c), you can dissolve 203.9 grams (or 200 grams) of sugar in 100 ml of water. as water temperature increases, the amount of sugar that will dissolve increases, up to about 500 grams of sugar in 100 ml water at 100 degrees c.

explanation:

the answer is b & c i believe : )