if you burned 3.0 g of magnesium and produced 5.0 g of product how much oxygen gas was used ?
2Mg + O2 ---> 2 MgO
Mg = 24.3 g/mol
O = 16 g/mol
so
48.6 grams Mg + 32 grams O2 --> 80.6 grams MgO
5/80.6 = .062 mol MgO produced
x/32 = .062
x = 1.985 grams O2
and we needed
y/48.6 = .062
y = 3.00 grams of Mg sure enough
To determine how much oxygen gas was used when 3.0 g of magnesium is burned and 5.0 g of product is produced, we will utilize the law of conservation of mass. According to this law, mass cannot be created or destroyed in a chemical reaction; it can only be rearranged.
The chemical equation for the reaction between magnesium and oxygen is:
2 Mg + O2 → 2 MgO
From the equation, we can see that 2 moles of magnesium react with 1 mole of oxygen gas to produce 2 moles of magnesium oxide.
To find the number of moles of magnesium in 3.0 g, we need to divide its mass by its molar mass. The molar mass of magnesium (Mg) is 24.31 g/mol. Thus, the number of moles of magnesium is calculated as follows:
moles of magnesium = mass of magnesium / molar mass of magnesium
moles of magnesium = 3.0 g / 24.31 g/mol = 0.1234 mol (rounded to four decimal places)
Since the stoichiometry of the reaction is 2 moles of magnesium react with 1 mole of oxygen gas, we know that 0.1234 moles of magnesium would react with:
moles of oxygen = moles of magnesium / stoichiometric coefficient
moles of oxygen = 0.1234 mol / 2 = 0.0617 mol (rounded to four decimal places)
Finally, to find the mass of oxygen gas used, we multiply the number of moles of oxygen by its molar mass. The molar mass of oxygen (O2) is 32.00 g/mol:
mass of oxygen = moles of oxygen * molar mass of oxygen
mass of oxygen = 0.0617 mol * 32.00 g/mol = 1.9744 g (rounded to four decimal places)
Therefore, approximately 1.9744 grams of oxygen gas was used in the reaction.