During combustion analysis, the unknown white powder in problem 2 was burned in the presence of oxygen gas to produce gaseous carbon dioxide and gaseous water. a) Write a complete, balanced chemical equation for this reaction. (Use the molecular formula of the unknown powder in your equation.) b) Using the balanced equation, calculate how much oxygen gas (in grams) is required to completely combust 38.43 grams of the white powder

a) 4C₆H₉O₉ + 15O₂ ---> 18H₂O + 24CO₂

b) 20.496 grams of oxygen is required to completely combust 38.43 grams of the white powder.

Note: The given molecular of the unknown white powder is C₆H₉O₉ according to the attachment below.

Explanation:

a) Balanced equation of reaction:

4C₆H₉O₉ + 15O₂ ---> 18H₂O + 24CO₂

b) molar mass of C₆H₉O₉ =225 g/mol; molar mass of O₂ = 32 g/mol

number of moles of C₆H₉O₉ in 38.43 g = 38.43/225 g/mol = 0.1708 moles

number of moles of oxygen required for combustion = 15/4 * 0.1708

number of moles of oxygen = 0.6405

mass of 0.6405 moles of oxygen = 0.6405 moles * 32 g/mol = 20.496 g

20.496 grams of oxygen is required to completely combust 38.43 grams of the white powder.

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