Metabolism of one mole of glucose, C6H12O6, releases 670 kcal. How much heat is released by the combustion of 0.300 moles of glucose?

To calculate the heat released by the combustion of 0.300 moles of glucose, we need to use the concept of stoichiometry.

The balanced equation for the combustion of glucose is:

C6H12O6 + 6 O2 -> 6 CO2 + 6 H2O

From the equation, we can see that 1 mole of glucose produces 6 moles of CO2 and 6 moles of H2O. The heat released per mole of glucose is given as 670 kcal.

To find the heat released by the combustion of 0.300 moles of glucose, we can use the following steps:

1. Calculate the heat released per mole of glucose:
Heat per mole of glucose = 670 kcal

2. Calculate the moles of glucose in the given sample:
Moles of glucose = 0.300 moles

3. Calculate the heat released by the combustion of 0.300 moles of glucose:
Heat released = Heat per mole of glucose x Moles of glucose
Heat released = 670 kcal/mole x 0.300 moles
Heat released = 201 kcal

Therefore, the combustion of 0.300 moles of glucose releases 201 kcal of heat.

To find the amount of heat released by the combustion of 0.300 moles of glucose, you need to use the concept of stoichiometry.

The given information states that the combustion of 1 mole of glucose releases 670 kcal. So, by using this information, we can set up a proportion to find the heat released by 0.300 moles of glucose.

Let's set up the proportion:

1 mole of glucose / 670 kcal = 0.300 moles of glucose / x kcal

To solve for x, which represents the heat released by the combustion of 0.300 moles of glucose, we can cross-multiply and solve for x:

1 mole of glucose * x kcal = 0.300 moles of glucose * 670 kcal

x = (0.300 moles of glucose * 670 kcal) / 1 mole of glucose

Now, let's substitute the values and calculate:

x = (0.300 * 670) / 1
x = 201 kcal

Therefore, the combustion of 0.300 moles of glucose releases 201 kcal of heat.

200