In the following reaction (already balanced) C3H8 + 5O2 —> 3CO2 + 4H2O How many grams of CO2 will be produced from 7.00 g of C3H8 and 98.0 g of O2? (Rounded to three significant figures)

Answer 1

First, we need to convert grams of each reactant to moles.

When we convert grams to moles or moles to grams, we use molar mass.

The molar mass of O₂ is 16.0g/mol, C₃H₈ is 44.1g/mol

-> O₂ : 98.0g / (16.0 g/mol) = 6.13 mol

-> C₃H₈ : 7.00g / (44.1g/mol) = 0.159 mol

Now, we need to find which one is limiting reactant.

From balanced equation, we get 1:5 ratio between C₃H₈ : O₂.

If we assume C₃H₈ is limiting reactant, we will consume 0.159 moles (all) and (0.159moles * 5 = 0.795)moles of O₂. Since we have enough amount of O₂ for this reaction, our assumption was correct.

Therefore, C₃H₈ is limiting reactant and O₂ is excess reactant.

Then, we take ratio between C₃H₈ : CO₂ to find the number of mole of CO₂ produced.

They have 1:3 ratio, so we will get (0.159 * 3 = 0.477)moles of CO₂.

For final answer, we need to convert moles to grams.

The molar mass of CO₂ is 44.01g/mol.

-> 0.477mol * 44.01g/mol = 21.0 g

Answer : 21.0g of CO₂

Hope this is helpful! :)