Question

Determine the mass in grams of C4H10 that is required to completely react to produce 8.70 mol of CO2 according to the following combustion reaction: 2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g).

Options:
Option 1: 660 g
Option 2: 760 g
Option 3: 860 g
Option 4: 960 g

Answer 1

To determine the mass of C4H10 required to produce 8.70 mol of CO2, we use stoichiometry. From the balanced equation, we find that 176.04 g of CO2 is produced by 2 moles of C4H10. Thus, the mass of C4H10 required is approximately 765 g.

Step-by-step explanation:

To determine the mass in grams of C4H10 required to produce 8.70 mol of CO2, we need to use stoichiometry. From the balanced equation, we can see that 2 moles of C4H10 produce 8 moles of CO2. So, for every 2 moles of C4H10, there are 8/2 = 4 moles of CO2. Using the molar mass of CO2 (44.01 g/mol), we can calculate the mass of CO2 produced: 4 mol CO2 * 44.01 g/mol = 176.04 g CO2.

Now, we need to calculate the mass of C4H10 that would produce 176.04 g CO2. From the balanced equation, we know that for every 2 moles of C4H10, we get 176.04 g CO2. So, for 1 mole of C4H10, we have 176.04 g CO2 / 2 mol C4H10 = 88.02 g CO2/mol C4H10.

Finally, to produce 8.70 mol of CO2, we multiply the mole ratio by the number of moles of CO2: 8.70 mol CO2 * (88.02 g CO2/mol C4H10) = 765.17 g C4H10. Therefore, the mass of C4H10 required is approximately 765 g (option 2).