if 6.47 moles of potassium chlorate is used in the reaction shown below, how many moles of C12H22O11 would also be used?

8KCLO3 (s) C12H22O11 (s) = 8KCl (s) + 12CO2 (g) + 11H2O(g)

Question

if 6.47 moles of potassium chlorate is used in the reaction shown below, how many moles of C12H22O11 would also be used?

8KCLO3 (s) C12H22O11 (s) => 8KCl (s) + 12CO2 (g) + 11H2O(g)

xishem · Answer

This is another mole-mole conversion.

Hint: For every 8 moles of KCLO3 that are consumed, how many moles of C12H22O11 will be consumed?

anonymous · Answer

1

xishem · Answer

Ok, so now you have a conversion factor to convert between moles of KCLO3 and C12H22O11 and vice-versa...$$\frac{1mol\ C_{12}H_{22}O_{11}}{8mol\ KCLO_3}=1$$

anonymous · Answer

ok so is it (6.47)(1/8)?

xishem · Answer

Yep (:. And you can always check yourself by making sure your units cancel. In this moles of KCLO3 cancel, and you are left with moles of sucrose.

anonymous · Answer

thanks again