Question

The molar mass of oxygen gas (O2) is 32.00 g/mol. The molar mass of C3H8 is 44.1 g/mol. What mass of O2, in grams, is required to completely react with 0.025 g C3H8

Answer 1

To determine the mass of O2 required to react with 0.025 g of C3H8, we calculate the moles of C3H8, use the 1:5 molar ratio with O2, and convert the moles of O2 required to grams, resulting in 0.09072 grams of O2 needed.

Step-by-step explanation:

The molar mass of oxygen gas (O2) is 32.00 g/mol and the molar mass of C3H8 is 44.1 g/mol. To find out how many grams of O2 are required to completely react with 0.025 g of C3H8, we need to use the stoichiometry of the combustion reaction.

The balanced chemical equation for the combustion of propane (C3H8) is: C3H8 + 5O2 → 3CO2 + 4H2O

From the equation, we see that one mole of C3H8 reacts with five moles of O2. Therefore, the mole ratio is 1:5.

First, we need to convert the given mass of C3H8 to moles:

0.025 g C3H8 x (1 mol C3H8 / 44.1 g C3H8) = 0.000567 moles of C3H8

Now, using the mole ratio, we calculate the moles of oxygen gas needed:

0.000567 moles C3H8 x (5 moles O2 / 1 mole C3H8) = 0.002835 moles O2

Finally, convert the moles of O2 to grams:

0.002835 moles O2 x (32.00 g O2 / 1 mol O2) = 0.09072 g O2

Therefore, 0.09072 grams of O2 is required to completely react with 0.025 g of C3H8.

Answer 2

The reaction formula of this is C3H8 + 5O2 --> 3CO2 + 4H2O. The ratio of mole number of C3H8 and O2 is 1:5. 0.025g equals to 0.025/44.1=0.00057 mole. So the mass of O2 is 0.00057*5*32=0.0912 g.