Question

2Mg(s)+O₂(g) ---> 2MgO(s)

what is the theoretical yield of MgO(s) in moles when 6.50g Mg(s) reacts with 3.45g of O2(g)​

Answer 1

Molar mass of Mg = 24.31 g/mol

mass of Mg = 6.5 g

we have below equation to be used:

number of mol of Mg,

n = mass of Mg/molar mass of Mg

=(6.5 g)/(24.31 g/mol)

= 0.2674 mol

Molar mass of O2 = 32 g/mol

mass of O2 = 3.45 g

we have below equation to be used:

number of mol of O2,

n = mass of O2/molar mass of O2

=(3.45 g)/(32 g/mol)

= 0.1078 mol

we have the Balanced chemical equation as:

2 Mg + O2 ---> 2 MgO

2 mol of Mg reacts with 1 mol of O2

for 0.2674 mol of Mg, 0.1337 mol of O2 is required

But we have 0.1078 mol of O2

so, O2 is limiting reagent

we will use O2 in further calculation

  

From balanced chemical reaction, we see that

when 1 mol of O2 reacts, 2 mol of MgO is formed

mol of MgO formed = (2/1)* moles of O2

= (2/1)*0.1078

= 0.216 mol

Answer: 0.216 mol