Question

For the reaction, calculate how many grams of the product form when 2.8 g of Mg completely reacts.

Assume that there is more than enough of the other reactant.
2Mg(s)+O2(g)-->2MgO(s)

Answer 1

4.64g MgO(s)

Step-by-step explanation:

Given 2Mg(s) + O₂(g) => 2MgO(s)

2.8g excess (?)

=> 2.8g/24.31 g/mol

= 0.115 mol. Mg(s) => 0.115 mol. MgO(s)

grams MgO(s) produced = 40.31g/mol. x 0.115 mol. = 4.64g (Theoritical Yield)