Carbon dioxide is a chemical compound composed of one carbon and two oxygen atoms. It is often referred to by its formula CO2. It is present in the Earth's atmosphere at a low concentration and acts as a greenhouse gas. In its solid state, it is called dry ice. It is a major component of the carbon cycle.
Answer:
d. 12.3 grams of Al2O3
Explanation:
The balanced chemical equation of this chemical reaction is as follows:
4Al + 3O2 --> 2Al2O3
Based on the balanced equation, 4 moles of aluminum (limiting reagent) reacts to form 2 moles of aluminum oxide (Al2O3).
First, we need to convert the mass of aluminum to moles using the formula;
mole = mass/molar mass
Molar mass of Al = 27g/mol
mole = 6.50/27
= 0.241mol of Al.
Hence, if 4 moles of aluminum (limiting reagent) reacts to form 2 moles of aluminum oxide (Al2O3).
Then, 0.241mol of Al will produce (0.241 × 2/4) = 0.241/2 = 0.121mol of Al2O3.
Convert this mole value to molar mass using mole = mass/molar mass
Molar mass of Al2O3 = 27(2) + 16(3)
= 54 + 48
= 102g/mol
mass = molar mass × mole
mass = 102 × 0.121
mass of Al2O3 = 12.34grams.
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Answer:
7.17 g/mL
Explanation:
Density = Mass / Volume
Given mass of object : 80.4g
We are not given the objects volume, however we are given that when put inside of a container with 20.0mL water, the level of the water rises to 31.2 mL.
31.2 - 20 = 11.2
The water level rose 11.2 mL when the object was placed inside of the cylinder containing the water therefore the object has a volume of 11.2mL
Now to find it's density
recall that density = mass / volume
mass = 80.4 and volume = 11.2mL
so density = 80.4/11.2 = 7.17 g/mL
Answer:
a. Oxygen is the limiting reagent. 
b. 
%
Explanation:
Hello,
a. Limiting reagent and sulfur trioxide's theoretical yield.
At first, we must compute the involved moles for both sulfur dioxide's and oxygen's as follows, considering the volumes in liters and the pressure in atm of 50.0mmHg*1atm/760mmHg=0.0658atm:
Afterwards, by considering the properly balanced chemical reaction:
We compute the oxygen's moles that completely reacts with the previously computed 
moles of 
as follows:
That result let us know that the oxygen is the limiting reagent since just 
moles are available in comparison with the 
moles that completely would react with moles of .
Now, to compute the theoretical yield of sulfur trioxide, we apply the following stoichiometric relationship:
b. Percent yield.
At first, we must compute the collected (real) moles of sulfur trioxide:
Finally, we compute the percent yield:
%
%
%
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