Consider the balanced equation: 2Mg(s) + O2(g) arrow 2MgO(s). 2mg(s) is 10.1g and 10.5g +O2 are missed i) determine the limiting reagent ii) determine the mass(in grams) of MgO produce iii) if the reaction produce 11.9g of MgO what is the percent yield?

Answer 1

i) To determine the limiting reagent, we need to calculate how much MgO would be produced from each reactant and see which one produces less MgO.

First, calculate the moles of each reactant:
- Moles of Mg = 10.1g / 24.3 g/mol = 0.415 moles
- Moles of O2 = 10.5g / 32 g/mol = 0.328 moles

Next, calculate how much MgO would be produced from each reactant:
- Mg (0.415 moles) could produce 0.415 moles of MgO
- O2 (0.328 moles) could produce 0.328 moles of MgO

Since the balanced equation is 2:1 for Mg to MgO, and for O2 to MgO, the limiting reagent is O2 since it produces less MgO (0.328 moles).

ii) To determine the mass of MgO produced, we need to calculate the mass of MgO produced from the limiting reagent:
- Moles of MgO from O2 = 0.328 moles
- Mass of MgO produced from O2 = 0.328 moles x 40.3 g/mol = 13.2g

Therefore, the mass of MgO produced from the reaction is 13.2g.

iii) To calculate the percent yield:
Actual yield = 11.9g
Theoretical yield (from O2) = 13.2g

Percent yield = (Actual yield / Theoretical yield) x 100
Percent yield = (11.9g / 13.2g) x 100
Percent yield = 0.9015 x 100
Percent yield = 90.15%

Therefore, the percent yield of the reaction is 90.15%.