Answer:
5.3g of MgO would be produced
Step-by-step explanation:
Based on the chemical reaction:
2Mg(s) + O₂(g) → 2MgO(s)
1 mole of Mg reacts per mole of oxygen.
To solve this question we need to convert the mass of each reactant to moles in order to find limiting reactant using the chemical equation:
Moles Mg -Molar mass: 24.305g/mol-:
3.8g Mg * (1mol / 24.305g) = 0.1563moles Mg
Moles O₂ -Molar mass: 32g/mol-:
2.1g O₂ * (1mol / 32g) = 0.0656 moles O₂
For a complete reaction of 0.0656 moles of O₂ are required:
0.0656 moles O₂ * (2 mol Mg / 1mol O₂) = 0.132 moles Mg
As there are 0.1563 moles of Mg, Mg is the excess reactant and O₂ the limiting reactant.
0.0656 moles O₂ produce:
0.0656 moles O₂ * (2 mol MgO / 1mol O₂) = 0.132 moles MgO
The mass is -Molar mass MgO: 40.3g/mol-:
0.132 moles MgO * (40.3g / mol) =
5.3g of MgO would be produced