Answer:
One gallon of octane produces approximately 7000 L of carbon dioxide.
Note:
I believe that the mass of octane should have been given as 2661 g. However, I understand that your instructor probably gave you this problem, so I will use 4000 g for the approximate mass of one gallon of octane. You can rework the problem on your own, substituting the correct masses of octane if you wish.
Step1. You must first determine the number of moles that are in 4000 g of octane, using the molar mass of octane. Step 2. Then you must determine the number of moles of carbon dioxide that can be produced by that number of moles of octane, based on the mole ratio between octane and carbon dioxide in the balanced equation. Step 3. Then use the ideal gas law to determine the volume in liters of carbon dioxide that can be formed.
Answer: 66.2 g
Explanation:
1) The ratio of Al in the molecule is 1 mol to 1 mol .
2) The mass of 1 mol of molecules of Al (CH2H3O2)3 is the molar mass of the compound.
3) You calculate the molar mass of the compound using the atomic masses of each atom, in this way:
Al: 27 g/mol
C: 2 * 3 * 12 g/mol = 72 g/mol
H: 3 * 3 * 1 g/mol = 9 g/mol
O: 2 * 3 * 16 g/mol = 96 g/mol
Molar mass = 27 g/mol + 72 g/mol + 9 g/mol + 96 g/mol = 204 g/mol
4) Set a proportion:
27 g/mol x
-------------------- = ----------
204 g/mol 500 g
5) Solve for x:
x = 500 g * 27 g/mol / 204 g/mol = 66.2 g
1. Double replacement (DR)
2. Decomposition (D)
<h3>Further explanation</h3>
1. Al2(SO4)3 + Ca3(PO4)2 -> 2AIPO4 + 3CaSO4
Double replacement (DR) : there is an ion exchange between two ion compounds in the reactant to form two new ion compounds in the product
General form :
AB + CD -> AD + CB
2. 2NaCIO3 → 2NaCl + 3O2
Decomposition (D) : Reactant breakdown into simpler ones(reverse of combination)
General form :
AB ---> A + B
Moles= mass\ relative formula mass(Ar)
moles of zinc= 7.9/30= 0.263
so we have 0.263 moles of zinc, and you need twice the amount of chlorine so therefore 0.526moles of chlorine= 0.526x 17=8.942g of chlorine
i cba to work the rest out but the most reasonable answer is 0.24 mol however if you need to use working outs, use the formula i provided earlier
