Answer
2229.32 grams
Step-by-step explanation
Given:
Volume of oxygen produced = 659 L
Pressure, P = 785 mmHg = 785/760 = 1.03 atm
Temperature, T = 30.0 °C = (30.0 +273.15 K) = 303.15 K
What to find:
The mass in grams of KClO3 that decompose.
Step-by-step solution:
The question asked to find the mass, in g, of KClO3 that decomposes to give a certain 659 L of O2.
Step 1: Write a balanced chemical equation for the reaction:
2KClO3 (s) ----> 2KCl (s) + 3O2 (g)
Step 2: Use the ideal gas equation to find the moles of O2 that form.
The ideal gas equation is:
Plugging in known values, and R = 0.082057 L.atm/mol.K we have:
Step 3: Use the coefficients of the chemical equation to find the relative number of moles of KClO3 that reacted:
From the balanced equation, 2 moles of KClO3 produced 3 moles of O2
Therefore, the moles of KClO3 that will produce 27.2866 moles of O2 is:
Finally, use the molar mass of potassium chlorate (122.55 g/mol) to find the number of grams that reacted:
2229.32 grams of KClO3 are needed to produce 659 L of oxygen at 785 mmHg and 30.0 °C.