Combustion of glucose takes place according to the equation $C_{6} H_{12} O_{6} + 6 O_{2} \to 6 C O_{2} + 6 H_{2} O; Δ H = - 72 k c a l / m o l$ .
How much energy will be released by the combustion of 1.6 g of glucose (Molecular mass of glucose = $180 g / m o l$ ) ?

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$0.064 k c a l$

$0.64 k c a l$

$6.4 k c a l$

$64 k c a l$

answer is B.

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Detailed Solution

Combustion of glucose takes place according to the equations $C_{6} H_{12} O_{6} + 6 O_{2} \to 6 C O_{2} + 6 H_{2} O; Δ H = - 72 k c a l$

$Δ H p e r 1.6 g = \frac{72 \times 1.6}{180} = 0.64 k c a l$

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