Answer:
this would be the .04 it is very simple
Explanation:
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: 1.52 atm
Explanation:
Given that:
Volume of gas V = 10.0L
Temperature T = 35.0°C
Convert Celsius to Kelvin
(35.0°C + 273 = 308K)
Pressure P = ?
Number of moles = 0.6 moles
Molar gas constant R is a constant with a value of 0.0821 atm L K-1 mol-1
Then, apply ideal gas equation
pV = nRT
p x 10.0L = 0.6 moles x (0.0821 atm L K-1 mol-1 x 308K)
p x 10.0L = 15.17 atm L
p = 15.17 atm L / 10.0L
p = 1.517 atm (round to the nearest hundredth as 1.52 atm)
Thus, the pressure of the gas is 1.52 atm
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