Question

Sucrose (C12H22011) is combusted in air according to the following reaction:

C12H22011(s) + O2(g) = CO2(g) + H2O(l )
How many moles of carbon dioxide would be produced by the complete combustion of 38.5 grams of sucrose in the presence of excess oxygen?

Answer 1

Okay, let's break this down step-by-step:

1) The molecular formula for sucrose is C12H22011. This means each mole of sucrose contains 12 moles of carbon and 22 moles of hydrogen.

2) You are combusting 38.5 grams of sucrose. To convert grams to moles, we divide by the molar mass:

Molar mass of C12H22011 = 342.3 g/mol

So 38.5 g / 342.3 g/mol = 0.113 moles of sucrose

3) According to the combustion reaction, each mole of sucrose produces 12 moles of CO2.

So 0.113 moles of sucrose will produce 0.113 * 12 = 1.356 moles of CO2.

4) Round to the nearest whole number:

1 mole of CO2

Therefore, the complete combustion of 38.5 grams of sucrose in excess oxygen would produce 1 mole of carbon dioxide.

Let me know if you have any other questions!