The answer is 60.3% magnesium, 39.7% oxygen.
Solution:
The chemical equation for the reaction is 2 Mg + O2 → 2 MgO.
Since magnesium reacts completely with oxygen, it is the limiting reactant in the reaction. Hence, we can use the number of moles of magnesium to get the mass of MgO produced:
moles of magnesium = 14.7g / 24.305g mol-1
= 0.6048 mol
mass of MgO = 0.6048mol Mg(2 mol MgO/2mol Mg)(40.3044g MgO/1 mol MgO)
= 24.376g MgO
We can now solve for the percentage of magnesium:
% Mg = (14.7g Mg / 24.376g MgO)*100% = 60.3%
We also use the number of moles of magnesium to get the mass of oxygen consumed in the reaction:
mass of O2 = 0.6048 mol Mg (1mol O2 / 2mol Mg) (31.998g / 1mol O2)
= 9.676g
The percentage of oxygen is therefore
% O2 = (9.676g O2 / 24.376g MgO)*100%
= 39.7%
Notice that we can just subtract the magnesium's percentage from 100% to get
% O2 = 100% - 60.3% = 39.7%
Answer:
P₂ = 5000 KPa
Explanation:
Given data:
Initial volume = 2.00 L
Initial pressure = 50.0 KPa
Final volume = 20.0 mL (20/1000=0.02 L)
Final pressure = ?
Solution:
The given problem will be solved through the Boly's law,
"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"
Mathematical expression:
P₁V₁ = P₂V₂
P₁ = Initial pressure
V₁ = initial volume
P₂ = final pressure
V₂ = final volume
Now we will put the values in formula,
P₁V₁ = P₂V₂
50.0 KPa × 2.00L = P₂ × 0.02 L
P₂ = 100 KPa. L/0.02 L
P₂ = 5000 KPa
Answer:
B
Explanation:
it will not be waterproof anymore
The maximum mass of B₄C that can be formed from 2.00 moles of boron (III) oxide is 55.25 grams.
<h3>What is the stoichiometry?</h3>
Stoichiometry of the reaction gives idea about the relative amount of moles of reactants and products present in the given chemical reaction.
Given chemical reaction is:
2B₂O₃ + 7C → B₄C + 6CO
From the stoichiometry of the reaction, it is clear that:
2 moles of B₂O₃ = produces 1 mole of B₄C
Now mass of B₄C will be calculated by using the below equation:
W = (n)(M), where
- n = moles = 1 mole
- M = molar mass = 55.25 g/mole
W = (1)(55.25) = 55.25 g
Hence required mass of B₄C is 55.25 grams.
To know more about stoichiometry, visit the below link:
brainly.com/question/25829169
#SPJ1
