Answer:
26.9 g
81%
Explanation:
The equation of the reaction is;
4 KO2(s) + 2 CO2(g) → 3 O2(g) + 2 K2CO3(s)
Number of moles of KO2= 27.9g/71.1 g/mol = 0.39 moles
4 moles of KO2 yields 2 moles of K2CO3
0.39 moles of KO2 yields 0.39 × 2/4 = 0.195 moles of K2CO3
Number of moles of CO2 = 57g/ 44.01 g/mol = 1.295 moles
2 moles of CO2 yields 2 moles of K2CO3
1.295 moles of CO2 yields 1.295 × 2/2 = 1.295 moles of K2CO3
Hence the limiting reactant is KO2
Theoretical yield = 0.195 moles of K2CO3 × 138.205 g/mol = 26.9 g
Percent yield = actual yield/theoretical yield × 100
Percent yield = 21.8/26.9 × 100
Percent yield = 81%
Answer:
Limiting: Lab covers. Books produced: 75. Excess amounts: In explanation
Explanation:
Amount that can be produced with just lab covers: 75 books (150/2)
Amount that can be produced with just lined paper: 150 books (7500/50)
Amount that can be produced with graph paper: 120 (3000/25)
Amount that can be produced with staples: Approximately 83 (250/3)
As we can see, the lab covers are limiting as they can only produce 75 books. So, we can only make 75 books.
You will have 3,750 lined paper left over (or 75 books)
(150-75=75 , 75*50=3,750)
You will have 1,125 graph paper left over (or 45 books)
(120-75=45 , 75*25=1,875 , 3000-1875=1125)
And then you will have approximately 25 staples left over (or 8 books)
(83-75=8 , 8*3=225, 250-225=25)
(Hopefully this is correct, I apologize if I messed up)
I still do not know but they do have quanza muscles
Here we have to get the correct molecular formula of the compound.
The molecular formula of the compound is C₆H₆O₆ i.e. option C.
Let assume, the empirical formula of the compound is 
.
The given molar mass of the compound is 176.124 g/mole.
The percent of carbon in the compound is 
×100 = 40.924.
The percent of hydrogen in the compound is 
×100 = 4.577.
The percent of oxygen in the compound is 
×100 = 54.498.
Now the ratio of the atomic number in the compound for carbon is 
= 3.410
Now the ratio of the atomic number in the compound for hydrogen is 
= 4.577
Now the ratio of the atomic number in the compound for oxygen is 
= 3.406
So, the C, H and O lowest ratio is 
= 1, 
= 1 and 
= 1
Thus the empirical formula of the compound is ![[CH_{1}O] _{n}](https://tex.z-dn.net/?f=%5BCH_%7B1%7DO%5D%20_%7Bn%7D)
(where n = integer.
12n + 1n + 16n = 176.24
29n = 176.24
n = 6 (approx)
Thus the molecular formula of the compound is C₆H₆O₆.
