Question

C12H22O11 + 12O2 ---> 12CO2 + 11H2O

If there are 10.0 g of sucrose and 8.0 g of oxygen, how many moles of sucrose are available for this reaction?

A.) 0.029 mol
B.) 0.250 mol
C.) 0.351 mol
D.) 3.00 mol

Answer 1

0.029 (A)

Step-by-step explanation:

Answer 2

Molar mass sucrose ( C12H22O11 ) = 342 g/mol

8.0 g of O2 is the limiting factor of the reaction reagent

C12H22O11 + 12 O2 ------------> 12 CO2 + 11 H2O

1 mol C12H22O11 ------------- 342 g
x mol C12H22O11 ------------- 10 g

x = 10 / 342

x = 0,029 mol of sucrose

answer A