Answer:- Third choice is correct, 17.6 moles
Solution:- The given balanced equation is:
Al_2(SO_4)_3+6KOH\rightarrow 2Al(OH)_3+3K_2SO_4
We are asked to calculate the moles of potassium hydroxide needed to completely react with 2.94 moles of aluminium sulfate.
From the balanced equation, there is 1:6 mol ratio between aluminium sulfate and potassium hydroxide.
It is a simple mole to mole conversion problem. We solve it using dimensional set up as:
2.94molAl_2(SO_4)_3(\frac{6molKOH}{1molAl_2(SO_4)_3})
= 17.6 mol KOH
So, Third choice is correct, 17.6 moles of potassium hydroxide are required to react with 2.94 moles of aluminium sulfate.
A. This is not a redox reaction. It is an example of combustion.
<h3>Combustion reaction of hydrocarbon</h3>
During the combustion of a hydrocarbon, the hydrocarbon reacts with oxygen to create carbon dioxide, water, and heat.
<h3>Example of combustion reaction</h3>
2C8H18 + 25O2 → 16CO2 + 18H2O
Thus, we can conclude that, this is not a redox reaction. It is an example of combustion.
Learn more about combustion here: brainly.com/question/9425444
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Answer:
water was added to powdered rock
Explanation:
Answer:
In He2 molecule,
Atomic orbitals available for making Molecular Orbitals are 1s from each Helium. And total number of electrons available are 4.
Molecular Orbitals thus formed are:€1s2€*1s2
It means 2 electrons are in bonding molecular orbitals and 2 are in antibonding molecular orbitals .
Bond Order =Electrons in bonding molecular orbitals - electrons in antibonding molecular orbitals /2
Bond Order =Nb-Na/2
Bond Order =2-2/2=0
Since the bond order is zero so that He2 molecule does not exist.
Explanation:
The balanced chemical reaction would be as follows:
<span>5P4O6 +8I2 ---> 4P2I4 +3P4O10
We are given the amount of reactants used for the reaction. We first need to determine the limiting reactant from the given amounts. We do as follows:
8.80 g P4O6 (1 mol / </span><span>219.88 g) = 0.04 mol P4O6
12.37 g I2 ( 1 mol / </span><span>253.809 g ) = 0.05 mol I2
Therefore, the limiting reactant is iodine since less it is being consumed completely in the reaction. We calculate the amount of P2I4 prepared as follows:
0.05 mol I2 ( 4 mol P2I4 / 8 mol I2 ) (</span><span>569.57 g / 1 mol) = 14.24 g P2I4</span>
