Answer:
6
Explanation:
Li has an oxydation number of 1 2*1 = 2
O has an oxidation number of -2 4(-2) = -8
S has an oxydation number of x 1 * x = x
The oxidation number on the molecule is 0.
So here is the equation
2 - 8 + x = 0 Combine like terms
-6 + x = 0 Add 6 to both sides
-6 + 6 + x = 6 Combine like terms
x = 6
Sulphur in this case has an oxidation number of 6
The product will not be affected by the addition of twice as much Na₂CO₃.
<h3>What is Limiting reagent in stoichiometry ?</h3>
- The maximum quantity of the end product determined by a balanced chemical equation is known as the Stoichiometry.
- The limiting reactant is the one that is consumed first and sets a limit on the quantity of product(s) that can be produced, and the one which remains unconsumed after the final reaction is in Excess.
- Calculate the moles of each reactant present and contrast it with the mole ratio of the reactants in the balanced equation to determine which reactant is the limiting one.
Here,taking the stoichiometry into consideration, we find that the reaction happens with 1:1 ratio; so, adding twice the amount of Na₂CO₃ will lead to its excess making the other the limiting reactant, hence, it would not affect the yield of the product.
To know more about the Limiting reactant, refer to:
brainly.com/question/14222359
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Carbon source is the answer for ur question
Please note that since magnesium is in group 2 it has a valency of 2+ and so the formula for magnesium hydroxide is
M
g
(
O
H
)
2
[Mistake in question]
This is a typical neutralization reaction of an acid with a base to form a salt and water. The reaction is exothermic, gives off heat,
Δ
H
<
0
, and may be balanced by adding balancing numbers in front, ie adding molecules, in order to ensure that the total number of atoms of each element is the same on the left and right hand sides of the equation.
Doing so we obtain :
2
H
3
P
O
4
+
3
M
g
(
O
H
)
2
→
M
g
3
(
P
O
4
)
2
+
6
H
2
O
