This problem is describing a reaction whereby aluminum oxide and iron metal are produced from iron (II) oxide and aluminum metal. Since the question is not specific, it is assumed the reaction should be written and also balanced as follows:
<h3>Chemical equations:</h3><h3 />
In chemistry, a chemical equation is used to represent a chemical reaction in which the reactants undergo a chemical change to yield specific products. Both sides are separated by a right-pointed arrow and the phase of each substance must be specified as well as balanced according to the law of conservation of mass, which demands the number of atoms to be the same before and after the reaction.
Here, in this problem, we can start by writing the reaction according to the given description:
![Al(s)+FeO(s)\rightarrow Al_2O_3(s)+Fe(s)](https://tex.z-dn.net/?f=Al%28s%29%2BFeO%28s%29%5Crightarrow%20Al_2O_3%28s%29%2BFe%28s%29)
However, it is not initially balanced because unequal number of atoms of aluminum and oxygen are present on both sides of the equation. In such a way, to overcome the aforementioned, we set a 2 on Al, a 3 on FeO and a 3 on Fe to obtain:
![2Al(s)+3FeO(s)\rightarrow Al_2O_3(s)+3Fe(s)](https://tex.z-dn.net/?f=2Al%28s%29%2B3FeO%28s%29%5Crightarrow%20Al_2O_3%28s%29%2B3Fe%28s%29)
Which now is abided by the law of conservation of mass.
Learn more about the law of conservation of mass: brainly.com/question/8062886
Answer :]
A.)Calculate the mass of ammonium sulfate that would be obtained by reacting with ammonia acid.
<em>Correct me if i'm wrong :]</em>
Answer: last one is right
Explanation: atoms of same element have always same number of protons but may have several isotopes which have different number of neutrons.
Answer:
12.32 L.
Explanation:
The following data were obtained from the question:
Mass of CH4 = 8.80 g
Volume of CH4 =?
Next, we shall determine the number of mole in 8.80 g of CH4. This can be obtained as follow:
Mass of CH4 = 8.80 g
Molar mass of CH4 = 12 + (1×4) = 12 + 4 = 16 g/mol
Mole of CH4 =?
Mole = mass/Molar mass
Mole of CH4 = 8.80 / 16
Mole of CH4 = 0.55 mole.
Finally, we shall determine the volume of the gas at stp as illustrated below:
1 mole of a gas occupies 22.4 L at stp.
Therefore, 0.55 mole of CH4 will occupy = 0.55 × 22.4 = 12.32 L.
Thus, 8.80 g of CH4 occupies 12.32 L at STP.