How many grams of oxygen gas must react to give 2.80g of ZnO?
2Zn(s)+O2(g)→2ZnO(s)
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To find the number of grams of oxygen gas required to produce 2.80g of ZnO, we need to use stoichiometry.
First, let's write down the balanced equation:
2Zn(s) + O2(g) -> 2ZnO(s)
From the equation, we can see that 2 moles of Zn react with 1 mole of O2 to produce 2 moles of ZnO.
To find the molar mass of ZnO, we add up the atomic masses of Zn and O:
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
Now, let's calculate the number of moles of ZnO:
moles of ZnO = mass of ZnO / molar mass of ZnO
moles of ZnO = 2.80g / 81.38 g/mol ≈ 0.034 mol
According to the balanced equation, 2 moles of ZnO are produced from 1 mole of O2. Therefore, we can say that for every 2 moles of ZnO, we need 1 mole of O2.
Now we can find the number of moles of O2 required:
moles of O2 = moles of ZnO / 2
moles of O2 = 0.034 mol / 2 = 0.017 mol
Finally, let's calculate the mass of O2 required:
mass of O2 = moles of O2 × molar mass of O2
mass of O2 = 0.017 mol × 32.00 g/mol (molar mass of O2)
mass of O2 ≈ 0.544 g
Therefore, approximately 0.544 grams of oxygen gas must react to give 2.80 grams of ZnO.