2Mg + O2--->2MgO

If more magnesium is added
a.the reaction will shift to the right.
b.the reaction will shift to the left.
c.there will be no shift.
and why im confused on why it would happen.

Is this an equilibrium reaction? If so, then we should write it as

Mg(s) + O2(g) ==> MgO(s)
Adding a solid will have no effect on the reaction at equilibrium.
If you are thinking of Le Chatelier's Principle, the easiest way to think of this is that a system at equilibrium will try to undo what we do to it. Therefore, for a system of A + B ==> C, adding A will cause the reaction to shift to the right. Why? because the reaction will try to undo what we did so it will shift so as to use up the added A and that means shifting to the right; i.e., reacting to use A to form C. B is will decrease, C will increase. I've use A, B, and C so we aren't dealing with solids.

To answer this question, we need to understand Le Chatelier's principle, which states that if a change is applied to a system in equilibrium, the system will respond in a way that minimizes the effect of that change. In this case, we have the reaction:

2Mg + O2 --> 2MgO

When more magnesium is added, the concentration of the reactant (Mg) increases. According to Le Chatelier's principle, the system will respond in a way that minimizes the effect of the increased concentration of magnesium.

In this reaction, magnesium is the reactant, and according to the balanced equation, it is being consumed in the forward direction to produce magnesium oxide (MgO). Therefore, to minimize the effect of the increased concentration of magnesium, the reaction will shift to the right, favoring the forward reaction.

Hence, the correct answer is:

a. The reaction will shift to the right.

Adding more magnesium will shift the equilibrium position to produce more magnesium oxide. This is because the increased concentration of magnesium promotes the formation of more magnesium oxide, which helps re-establish the equilibrium in the system.

It's important to note that Le Chatelier's principle is only applicable to reactions in dynamic equilibrium and not to reactions that have not yet reached equilibrium.