2 Mg + O2 --> 2 MgO If 6.9 g of Mg react with 450 g of O2 what mass of product forms?

Question

Im wondering how I should do this

Answer 1

In order to answer this, we need to first find out which reactant is in limiting supply. That reactant will dictate how much product can be formed.

For Mg: 6.9 g Mg x 1 mol Mg/24.3 g = 0.284 mol Mg x 2 mol MgO/2 mol Mg = 0.283mol MgO

For O2: 450 g O2 x 1 mol O2/32 g = 14.1 mol O2 x 2 mol MgO / 1 mol O2 = 28.2 mol MgO

Clearly the Mg is in limiting supply, so only 0.283 moles of MgO can be produced

The mass of MgO is calculated as...

0.283 mol MgO x 40.3 g/mol = 11.4 g MgO can theoretically be formed