Question

question 33 (1 point)
consider respiration, one of the most common chemical reactions on earth. what is
the limiting reagent in the reaction of 25 grams of glucose with 40 grams of oxygen?
c₆h₁₂o₆ + o₂ → co₂ + h₂o
o₂
co₂
h₂
c₆h₁₂o₆
h₂o

Explanation:

Step1: Balance the chemical equation

The balanced equation for glucose respiration is \(C_{6}H_{12}O_{6}+6O_{2}\to6CO_{2}+6H_{2}O\)

Step2: Calculate the molar mass

• Molar mass of \(C_{6}H_{12}O_{6}\): \(M_{glucose}=(6\times12)+(12\times1)+(6\times16)= 180\space g/mol\)
• Molar mass of \(O_{2}\): \(M_{O_{2}} = 32\space g/mol\)

Step3: Calculate the number of moles

• Moles of glucose: \(n_{glucose}=\frac{25\space g}{180\space g/mol}\approx0.139\space mol\)
• Moles of \(O_{2}\): \(n_{O_{2}}=\frac{40\space g}{32\space g/mol} = 1.25\space mol\)

Step4: Use the mole - ratio from the balanced equation

From \(C_{6}H_{12}O_{6}+6O_{2}\to6CO_{2}+6H_{2}O\), the mole ratio of \(C_{6}H_{12}O_{6}\) to \(O_{2}\) is \(1:6\)

• If all \(0.139\space mol\) of \(C_{6}H_{12}O_{6}\) reacts, moles of \(O_{2}\) needed \(n_{O_{2}\text{(needed)}}=0.139\times6 = 0.834\space mol\)
• If all \(1.25\space mol\) of \(O_{2}\) reacts, moles of \(C_{6}H_{12}O_{6}\) needed \(n_{glucose\text{(needed)}}=\frac{1.25}{6}\approx0.208\space mol\)

Since we have only \(0.139\space mol\) of \(C_{6}H_{12}O_{6}\) (less than \(0.208\space mol\) needed if \(O_{2}\) is to react completely), and \(O_{2}\) available (\(1.25\space mol\)) is more than \(O_{2}\) needed (\(0.834\space mol\)) for complete reaction of glucose.

Answer:

\(C_{6}H_{12}O_{6}\)