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Chapter 9: Problem 17

How many grams of oxygen does it take to burn \(10.0 \mathrm{~g}\) of glucose? Answer:

Short Answer

Expert verified
It would require approximately \(10.7~\mathrm{g}\) of oxygen to burn \(10.0~\mathrm{g}\) of glucose.

Step by step solution

01

Write the balanced chemical equation for glucose combustion

The balanced chemical equation for the combustion of glucose (\(C_6H_{12}O_6\)) is as follows: \[C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O\] This tells us that one mole of glucose reacts with six moles of oxygen gas (\(O_2\)) to produce 6 moles of carbon dioxide gas (\(CO_2\)) and 6 moles of water.
02

Convert grams of glucose to moles of glucose

First, we need to determine the number of moles of glucose in 10.0 grams. To do this, we usually divide the mass of the given substance by its molar mass: Moles \(C_6H_{12}O_6 = \frac{mass}{molar~mass}\) The molar mass of glucose is approximately: \(C_6H_{12}O_6 = (6 \times 12.01) + (12 \times 1.01) + (6 \times 16.00) = 180.18~\mathrm{g/mol}\) Now, we can determine the number of moles of glucose: Moles of \(C_6H_{12}O_6 = \frac{10.0~\mathrm{g}}{180.18~\mathrm{g/mol}} \approx 0.0555~ \mathrm{mol}\)
03

Determine moles of oxygen required using stoichiometry

From the balanced equation, we know the following relationship between moles of glucose and moles of oxygen: 1 mole \(C_6H_{12}O_6 : 6 ~\mathrm{mol}~ O_2\) So, using the stoichiometric coefficients, we can compute the moles of oxygen needed: Moles of \(O_2 = moles~of~C_6H_{12}O_6 \times 6\) Moles of \(O_2 \approx 0.0555~ \mathrm{mol} \times 6 = 0.333~\mathrm{mol}\)
04

Convert moles of oxygen needed to grams of oxygen

Now, we need to convert the moles of oxygen needed to grams. \(Mass ~of~ O_2 = moles ~\times molar ~mass\) The molar mass of one mole of oxygen gas is approximately 32.00 g/mol (since one molecule of \(O_2\) contains two oxygen atoms): \(Mass ~of~ O_2 \approx 0.333~\mathrm{mol} \times 32.00~\mathrm{g/mol} \approx 10.7~g\) Thus, it would require approximately 10.7 grams of oxygen to burn 10.0 grams of glucose. Answer: \boxed{10.7~\mathrm{g}}.

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