How many grams of oxygen gas must react to give 2.70g of ZnO?
2Zn(s)+O2(g)→2ZnO(s)
To determine the number of grams of oxygen gas needed to react with 2.70 grams of ZnO, you can use the stoichiometry of the balanced chemical equation. Here's how to do it:
Step 1: Determine the molar mass of ZnO:
The molar mass of ZnO is calculated by summing the atomic masses of its constituent elements:
Zn: 65.38 g/mol
O: 16.00 g/mol
Molar mass of ZnO = 65.38 g/mol + 16.00 g/mol = 81.38 g/mol
Step 2: Convert the mass of ZnO to moles:
To convert grams of ZnO to moles, divide the mass by the molar mass:
2.70 g ZnO / 81.38 g/mol = 0.0332 mol ZnO
Step 3: Use the stoichiometry of the balanced equation to determine the moles of oxygen gas:
According to the balanced equation, the stoichiometric ratio between ZnO and O2 is 2:1. This means that for every 2 moles of ZnO, 1 mole of O2 is needed.
0.0332 mol ZnO * (1 mol O2 / 2 mol ZnO) = 0.0166 mol O2
Step 4: Convert the moles of oxygen gas to grams:
To convert moles of oxygen gas to grams, multiply the moles by the molar mass of O2:
0.0166 mol O2 * 32.00 g/mol = 0.532 g O2
Therefore, 0.532 grams of oxygen gas must react to give 2.70 grams of ZnO.
mols ZnO desired = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols ZnO to mols O2.
Then convert mols O2 to grams. g = mols x molar mass = ?