Answer:
%age Yield = 34.21 %
Explanation:
The balance chemical equation for the decomposition of KClO₃ is as follow;
3 KOH + H₃PO₄ → K₃PO₄ + 3 H₂O
Step 1: Calculate moles of H₃PO₄ as;
Moles = Mass / M/Mass
Moles = 334.6 g / 97.99 g/mol
Moles = 3.414 moles
Step 2: Find moles of K₃PO₄ as;
According to equation,
1 moles of H₃PO₄ produces = 1 moles of K₃PO₄
So,
3.414 moles of H₃PO₄ will produce = X moles of K₃PO₄
Solving for X,
X = 1 mol × 3.414 mol / 1 mol
X = 3.414 mol of K₃PO₄
Step 3: Calculate Theoretical yield of K₃PO₄ as,
Mass = Moles × M.Mass
Mass = 3.414 mol × 212.26 g/mol
Mass = 724.79 g of K₃PO₄
Also,
%age Yield = Actual Yield / Theoretical Yield × 100
%age Yield = 248 g / 724.79 × 100
%age Yield = 34.21 %
The number of moles of oxygen required to generate 28 moles of water from the reaction is 14 moles
<h3>Balanced equation </h3>
2H₂ + O₂ —> 2H₂O
From the balanced equation above,
2 moles of water were obtained from 1 mole of oxygen
<h3>How to determine the mole of oxygen needed </h3>
From the balanced equation above,
2 moles of water were obtained from 1 mole of oxygen
Therefore,
28 moles of water will be obtained from = 28 / 2 = 14 moles of oxygen
Thus, 14 moles of oxygen are needed for the reaction
Learn more about stoichiometry:
brainly.com/question/14735801
Answer:
3 moles of O2 needed
Explanation:
2 moles of Mg to one mole O2
so 3 moles of O2 needed
The much of the sample that would remain unchanged after 140 seconds is 2.813 g
Explanation
Half life is time taken for the quantity to reduce to half its original value.
if the half life for Scandium is 35 sec, then the number of half life in 140 seconds
=140 sec/ 35 s = 4 half life
Therefore 45 g after first half life = 45 x1/2 =22.5 g
22.5 g after second half life = 22.5 x 1/2 =11.25 g
11.25 g after third half life = 11.25 x 1/2 = 5.625 g
5.625 after fourth half life = 5.625 x 1/2 = 2.813
therefore 2.813 g of Scandium 47 remains unchanged.
For this question you need to find out all of the stages of a star and the process of a nuclear fusion. Then compare what's different about it and what's the same.
