Answer
Approximately 59.68 Cal.
Work Step by Step
1. Cover the mass of $C_6H_{12}O_6$ to nº of moles:
$mm (C_6H_{12}O_6) = 12*6 + 1*12 + 16*6 = 72 + 12 + 96 = 180$
$n(moles) = \frac{mass(g)}{mm}$
$n(moles) = \frac{16}{180} \approx 0.0888$
2. Now, convert that number to kJ:
$0.0888mol(C_6H_{12}O_6) \times \frac{2812}{1mol(C_6H_{12}O_6)}$
$249.7kJ$
3. Finally, convert that number to calories:
$249.7kJ \times \frac{1 Cal}{4.184kJ}$
$\approx 59.68$ $Cal$
