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) The problem is that you get a salty taste in our mouth when you drink orange juice after brushing our teeth.
Explanation:
Answer:
Moles of H₂S needed = 6.2 mol
Moles of SO₂ produced = 6.2 mol
Explanation:
Given data:
Number of moles of O₂ = 9.3 mol
Moles of H₂S needed = ?
Moles of SO₂ produced = ?
Solution:
Chemical equation:
2H₂S + 3O₂ → 2SO₂ + 2H₂O
Now we will compare the moles of oxygen with H₂S.
O₂ : H₂S
3 : 2
9.3 : 2/3×9.3 = 6.2 mol
Now we will compare the moles of SO₂ with both reactant.
O₂ : SO₂
3 : 2
9.3 : 2/3×9.3 = 6.2 mol
H₂S : SO₂
2 : 2
6.2 : 6.2 mol
So 6.2 moles of SO₂ are produced.