Suppose 1.043 g of magnesium is heated in air. What is the theoretical amount of magnesium oxide that should be produced?

Question

Wyatt Atkins · Accepted Answer

See explanation.

When magnesium is burned, the following happens: #2Mg(s) + O_2(g) -> 2MgO(s)# To calculate the amount of magnesium oxide that will be produced when #1.043g# are burnt: #?gMgO=underbrace(1.043cancel(color(blue)(gMg))xx(1cancel(color(red)(molMg)))/(24.31cancel(color(blue)(gMg))))_("convert g to mol")xxunderbrace((2cancel(color(green)(molMgO)))/(2cancel(color(red)(molMg))))_("Molar ratio")xxunderbrace((40.31gMgO)/(1cancel(color(green)(molMgO))))_("convert mol to g")=1.729gMgO#

Cooper Anderson · Answer

undefined The molar mass of magnesium (Mg) is approximately 24.31 g/mol, and the molar mass of oxygen (O) is approximately 16.00 g/mol.

To find the theoretical amount of magnesium oxide (MgO) produced, we first need to determine the moles of magnesium used. Then, we use the mole ratio between magnesium and oxygen in magnesium oxide to find the moles of oxygen, and finally, convert that to grams of magnesium oxide.

1. Convert the mass of magnesium to moles:
$$ \text{moles of Mg} = \frac{\text{mass of Mg}}{\text{molar mass of Mg}} $$

2. Use the mole ratio between Mg and O in MgO (1:1) to find the moles of oxygen:
$$ \text{moles of O} = \text{moles of Mg} $$

3. Convert the moles of oxygen to grams of MgO:
$$ \text{mass of MgO} = \text{moles of O} \times \text{molar mass of MgO} $$

Substituting the given values:
$$ \text{mass of MgO} = \left( \frac{1.043 \, \text{g}}{24.31 \, \text{g/mol}} \right) \times \left( 1 \, \text{mol} \right) \times \left( \frac{16.00 \, \text{g}}{\text{mol}} \right) $$

$$ \text{mass of MgO} = 0.04284 \, \text{g} $$