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Chapter 1: Problem 18

Select the correct alternative from the given choices. When \(180 \mathrm{~g}\) of glucose is subjected to combustion, the volume of \(\mathrm{CO}_{2}\) liberated at \(\mathrm{STP}\) is (a) \(22.4 \ell\) (b) \(67.2\) \ell (c) \(44 \ell\) (d) \(134.4 \ell\)

Short Answer

Expert verified
Answer: The volume of CO₂ produced is 134.4 L.

Step by step solution

01

Write the balanced chemical equation for the combustion of glucose

The combustion of glucose can be represented by the following equation: C₆H₁₂O₆ (s) + 6O₂ (g) → 6CO₂ (g) + 6H₂O (l) This equation needs to be balanced, and it already is balanced (there's already an equal number of atoms of each type on both sides).
02

Calculate the moles of glucose and CO₂

Given the mass of glucose as 180g, we can calculate the number of moles of glucose using its molar mass (C=12, H=1, O=16). Molar mass of glucose = 6 x 12 + 12 x 1 + 6 x 16 = 180g/mol Number of moles of glucose = mass / molar mass n(glucose) = 180g / 180g/mol = 1 mol According to the balanced chemical equation, 1 mole of glucose produces 6 moles of CO₂, so: n(CO₂) = 6 x n(glucose) n(CO₂) = 6 x 1 = 6 moles
03

Calculate the volume of CO₂ at STP

We know that at STP (Standard Temperature and Pressure, which is 0 °C and 1 atm), 1 mole of any gas occupies 22.4 L. Hence, we can find the volume of CO₂ by multiplying the number of moles of CO₂ by the volume occupied by 1 mole of gas at STP. Volume of CO₂ = n(CO₂) x Volume of 1 mole of gas at STP Volume of CO₂ = 6 moles x 22.4 L/mol = 134.4 L The correct answer is (d) \(134.4 \ell\).

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Most popular questions from this chapter

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