The balanced chemical equation for the combustion of propane is

C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)
C3 H 8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

Which statement is correct about the complete combustion of 3.00 mole of propane, C3H8?

a. 12.00 mol H2O are produced.
b. 3.00 g CO2 are produced.
c. 3.00 mol CO2 are produced.
d. 12.00 g H2O are produced.

Option a. 12.00 mol H2O are produced.

Explanation:

C3H8 + 5O2 → 3CO2 + 4H2O

From equation,

1mole of C3H8 produce 4moles of H2O.

Therefore, 3moles of C3H8 will produce = 3 x 4 = 12moles of H2O

Option A is correct 12.00 mol H2O are produced.

Explanation:

Step 1: Data given

Number of moles of propane = 3.00 moles

Step 2: The balanced equation

C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)

Step 3: Calculate moles

For 1 mol propane we need 5 moles O2 to produce 3 moles of CO2 and 4 moles H2O

For 3.00 moles propane we need:

5*3.00 = 15.00 moles O2

3*3.00 = 9.00 moles CO2

4*3.00 = 12.00 moles H2O

Step 4: Calculate mass

Mass = moles * molar mass

Mass H2O = 12.00 mol * 18.02 g/mol

Mass H2O = 216.2 grams

Mass CO2 = 9.00 moles * 44.01 g/mol

Mass CO2 = 396.1 grams

Option A is correct 12.00 mol H2O are produced.

447.63 g HClO₃

Explanation:Moles are calculated by dividing the mass of a compound by its relative formula mass.Therefore, when given the number of moles, to get the mass of the compound we multiply the number moles by the relative formula mass of the compound.Relative formula mass i given by adding the atomic masses of elements present in the compound.

In this case;

Relative formula mass of HClO₃ = 84.459 g

We are given, Moles of HClO₃ = 5.3 moles

Therefore;

Mass of HClO₃ = 5.3 moles × 84.459 g/mol

                        = 447.6327 g

                       = 447.63 g (2 d.p.)

Thus, the mass of HClO₃ in 5.3 moles is 447.63 g