The formation reaction of magnesium oxide is as follows:
Mg(s) + 0.5O2(g) → MgO(s)
This reaction can be constructed from the following reactions.
In the blanks before each reaction, add a coefficient that shows how they must be scaled before being added to give the reaction above. If the reversal of a reaction is required, indicate this using a negative sign before the coefficient. For example, if a reaction must be multiplied by 2 and reversed, write "–2."

(a) ___ Mg(s) + 2HCl(aq) → Mg2+(aq) + 2Cl–(aq) + H2(g)
(b) ___ MgO(s) + 2HCl(aq) → Mg2+(aq) + 2Cl–(aq) + H2O(l)
(c) ___ H2(g) + 0.5O2(g) → H2O(l)

(a) 1

(b) -1

(c) 1

Explanation:

In this type of questions use the strategy to look for placing the products and reactants on their right side.

For b) we notice MgO is a product in our reaction, therefore reverse the equation, the coefficient is 1 since there is one mol MgO product

Mg2+(aq) + 2Cl–(aq) + H2O(l) →  MgO(s) + 2HCl(aq)

We can notice that HCl is not a product in our final reaction, so we want to cancel it. If we add (a) to the one above, we get:

Mg(s) + 2HCl(aq) → Mg2+(aq) + 2Cl–(aq) + H2(g)

                                    +

Mg2+(aq) + 2Cl–(aq) + H2O(l) →  MgO(s) + 2HCl(aq)

Mg(s)        +           H2O(l)           →      H2(g) + MgO(s)

Finally adding this last equation to (c) we get our sought equation:

Mg(s)        +           H2O(l)           →      H2(g) + MgO(s)

                                         +

H2(g)            +     0.5O2(g)        H2O(l)

Mg(s)          +         0.5O2(g) →         MgO(s)

So we only needed to reverse the sign for the first equation, and our answer should be

(a) 1

(b) -1

(c) 1

the answer is barium and copper bromide