Combustion of glucose takes place according to the equation
$C_{6} H_{12} O_{6} + 6 O_{2} \to 6 C O_{2} + 6 H_{2} O; Δ H = - 72 k c a l$
$Δ H$ per $1.6 g = \frac{72 \times 1.6}{180} = 0.64 k c a l$

Question

Combustion of glucose takes place according to the equation C 6 H 12 O 6+6 O 2 arrow 6 CO 2+6 H 2 O ; Δ H =-72 kcal / mo How much energy will be released by the combustion of 1.6 g of glucose (Molecular mass of glucose =180 g / mol )? (A) 0.064 kcal (B) 0.64 kcal (C) 6.4 kcal (D) 64 kcal

Tardigrade Admin · Accepted Answer

Combustion of glucose takes place according to the equation C 6 H 12 O 6+6 O 2 arrow 6 CO 2+6 H 2 O ; Δ H =-72 kcal Δ H per 1.6 g =(72× 1.6/180)=0.64 kcal

Q. Combustion of glucose takes place according to the equation C6​H12​O6​+6O2​→6CO2​+6H2​O;ΔH=−72kcal/mo How much energy will be released by the combustion of 1.6g of glucose (Molecular mass of glucose =180g/mol )?

0.064 kcal

0.64 kcal

6.4 kcal

64 kcal

Combustion of glucose takes place according to the equation C6​H12​O6​+6O2​→6CO2​+6H2​O;ΔH=−72kcal ΔH per 1.6g=18072×1.6​=0.64kcal

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

Combustion of glucose takes place according to the equation
$C_{6} H_{12} O_{6} + 6 O_{2} \to 6 C O_{2} + 6 H_{2} O; Δ H = - 72 k c a l$
$Δ H$ per $1.6 g = \frac{72 \times 1.6}{180} = 0.64 k c a l$