Step 1
The reaction is written and balanced:
4 Rb + O2 =>2 Rb2O
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Step 2
Define % yield of product (Rb2O) = (Actual yield/Theoretical yield) x 100
The actual yield is provided by the exercise = 39.7 g
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Step 3
Determine the limiting reactant. The molar masses are needed to solve this:
For Rb) 85.4 g/mol
For O2) 32 g/mol
Procedure:
4 Rb + O2 =>2 Rb2O
4 x 85.4 g Rb ----- 32 g O2
82.4 g Rb ----- X = 7.72 g O2 are needed
For 82.4 g Rb, 7.72 g O2 is needed, but there is 11.6 g O2. Therefore, O2 is the excess agent. Rb is the limiting reactant.
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Step 4
Determine the theoretical yield from the limiting reactant:
The molar mass Rb2O) 187 g/mol
Procedure:
4 x 85.4 g Rb ------ 2 x 187 g Rb2O
82.4 g Rb ------ X = 90.2 g Rb2O = Theoretical yield
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Step 5
% yield = Actual y./Theoretical y. x 100 = (39.7 g/90.2 g) x 100 = 44 % approx.
Answer: % yield = 44 %
Answer:
114 K
Explanation:
Given data
- Volume of oxygen (V): 629 mL = 0.629 L
- Pressure of oxygen (P): 0.500 atm
- Moles of oxygen (n): 0.0337 mol
We can calculate the temperature at which the student collected the oxygen using the ideal gas equation.
The oxygen gas was collected at 114 K.
Spinning top, moving car, and rolling ball have kinetic energy I believe
