Question

Pure magnesium metal is often found as ribbons and can easily burn in the presence of oxygen. When 3.28 g of magnesium ribbon burns with 7.25 g of oxygen, a bright, white light and a white, powdery product are formed. Enter the balanced chemical equation for this reaction. Be sure to include all physical states. equation: What is the limiting reactant? magnesium oxygen The reaction goes to completion, but in the process of recovering the product, some of it was lost. The the percent yield for the reaction is 83.7%. How many grams of product are recovered? mass of product recovered: g How many grams of the excess reactant remain? Assume the reaction goes to completion. mass of excess reactant: g

Answer 1

2 Mg(s) + O2(g)→ 2 MgO(s)

Step-by-step explanation:

Number of moles of magnesium reacted = mass/ molar mass = 3.28g/24gmol-1 = 0.137 moles

From the balanced reaction equation;

2 moles of Mg yields 2 moles of MgO

0.137 moles of Mg yields 0.137 moles of MgO

For oxygen;

Number of moles of oxygen reacted = mass/ molar mass = 7.5g/32gmol-1 = 0.234 moles of O2

If 2_moles of oxygen yields 2 moles of MgO

0.234 moles of oxygen yields 0.234 moles of MgO

Hence Mg is the limiting reactant

Theoretical yield of MgO = 0.137 moles × 40 gmol-1 = 5.48 g of MgO

% yield of MgO = 83.7%

But

% yield = actual yield/theoretical yield × 100

actual yield = % yield × theoretical yield / 100

actual yield= 83.7 × 5.48/100

Actual yield = 4.59 g

According to the equation;

2 moles of Mg reacts with 2 moles of oxygen

Hence 0.137 moles of Mg will react with 0.137 moles of O2

Amount of excess reactant = 0.234 moles - 0.137 moles = 0.097 moles

Mass of excess reactant = 0.097 moles × 32gmol-1 = 3.1g of O2