Suppose two chemical reactions are linked together in a way that the O2 produced inthe first reaction goes on to react completely with Mg to form MgO in the second reaction. Reaction one: 2 KClO3 3 O2→ + 2 KClReaction two: 2 Mg + O2 2 MgO→If you start with 4 moles of KClO3, how many moles of MgO could eventually form?

A) 2 moles
B) 4 moles
C) 6 moles
D) 12 moles
E) none of the above

12 moles of MgO could eventually form.

Explanation:

In the reaction:

2KClO₃ → 3O₂ + 2KCl

2 moles of KClO₃ produce 3 moles of O₂. 4 moles of KClO₃ produce:

4 mol KClO₃ × (3 mol O₂ / 2 mol KClO₃) = 6 mol O₂

In the second reaction:

2Mg + O₂ → 2MgO

1 mole of O₂ produce 2 moles of MgO. Thus, 6 mol of O₂ produce:

6 mol O₂ × (2 mol MgO  / 1 mol O₂) = 12mol MgO

Thus, 12 moles of MgO could eventually form.

the answer for sodium is 0.186

the answer for rubidium is 0.251