<h2>
Answer: 131.9 g</h2>
<h3>
Explanation:</h3>
<u>Write a Balanced Equation for the decomposition</u>
CaCO₃ → CaO + CO₂
<u></u>
<u>Find Moles of CO₂ Produced</u>
Since the mole ratio of CaCO₃ to CO₂ is 1 to 1,
the moles of CaCO₃ = moles of CO₂
moles of CaCO₃ = mass ÷ molar mass
= 300 g ÷ 100.087 g/mol
= 2.997 moles
∴ moles of CO₂ = 2.997 moles
<u>Determine Mass of CO₂</u>
Mass = moles × molar mass
= 2.997 mol × 44.01 g/mol
= 131.9 g
<u></u>
<h3>∴ when 300 g of calcium carbonate is decomposed, it produces 131.9 g of carbon dioxide.</h3>
<span>1.
London forces. 4. dipole - dipole. Due to a small hydrogen atom and a
much large fluorine atom, with a large , positive nuclus and large,
negative, p orbitals . This makes it very polar.</span>
We see from the chemical formula itself that there is 1
mole of F2 for every 1 mole of CaF2, hence the number of moles of CaF2 is also:
moles CaF2 = 1.23 moles
The molar mass of CaF2 is 78.07 g/mol, so the mass is:
mass CaF2 = 78.07 g / mol * 1.23 mol
<span>mass CaF2 = 96.03 grams</span>
Answer: Percent yield represents the ratio between what is experimentally obtained and what is theoretically calculated, multiplied by 100%.
% yield
=
actual yield
theoretical yield
⋅
100
%
So, let's say you want to do an experiment in the lab. You want to measure how much water is produced when 12.0 g of glucose (
C
6
H
12
O
6
) is burned with enough oxygen.
C
6
H
12
O
6
+
6
O
2
→
6
C
O
2
+
6
H
2
O
Since you have a
1
:
6
mole ratio between glucose and water, you can determine how much water you would get by
12.0
g glucose
⋅
1 mole glucose
180.0 g
⋅
6 moles of water
1 mole glucose
⋅
18.0 g
1 mole water
=
7.20
g
This represents your theoretical yield. If the percent yield is 100%, the actual yield will be equal to the theoretical yield. However, after you do the experiment you discover that only 6.50 g of water were produced.
Since less than what was calculated was actually produced, it means that the reaction's percent yield must be smaller than 100%. This is confirmed by
% yield
=
6.50 g
7.20 g
⋅
100
%
=
90.3
%
You can backtrack from here and find out how much glucose reacted
65.0 g of water
⋅
1 mole
18.0 g
⋅
1 mole glucose
6 moles water
⋅
180.0 g
1 mole glucose
=
10.8
g
So not all the glucose reacted, which means that oxygen was not sufficient for the reaction - it acted as a limiting reagent.
Explanation:
