Instructions: Show all work/calculations for each problem to get credit, This as
ID: 591623 • Letter: I
Question
Instructions: Show all work/calculations for each problem to get credit, This assignment is worth 14 points and due on exam day. You must show all work. 1. Using the information below, calculate the amount of water that can be produced by the reaction of 10 grams of potassium chlorate with 4 grams of table sugar. Circle the limiting reagent. (2 pts) KClo, + C12H22011 KCl + CO2 + H2O 2. From the same reaction above, what volume of CO, gas can be produced at room temperature and 720 mm Hg? (2 pt) 3. Solid potassium chlorate can be decomposed to potassium chloride and oxygen by heating according to the balanced equation below. What mass in grams of potassium chlorate is required to produce 5.0 L of O2 gas at 20°C and 700 mm Hg? (2 pts) 2 KCIO3 (s) 2 KCl (s) + 32 (g)Explanation / Answer
1.
The balanced equation is,
8 KClO3 + C12H22O11 -------------> 8 KCl + 12 CO2 + 11 H2O
molar mass of KClO3 = 39 + 35.5 + 48 = 122.5 g/mol
molar mass of sucrose = 342 g/mol
From the balanced equation,
8 mol of KClO3 needs 1 mol of Sucrose
8 * 122.5 g. of KClO3 needs 342 g. of sucrose
then,
10 g. of KClO3 needs 10 * 342 / (8*122.5) = 3.49 g. of sucrose
But we have 4 g. of sucrose.
Hence KClO3 is limiting reagent.
from the balanced equation,
8 *122.5 g.of KClO3 forms 11 * 18 g. of H2O
then, 10 g. of KClO3 forms 10 * 11 * 18 / (8*122.5) = 2.02 g. of H2O