How many grams of oxygen gas must react to give 2.70 g of ZnO?
2Zn(s)+O2(g)→2ZnO(s)
Express your answer with the appro
K=FC^2
To determine the number of grams of oxygen gas needed to produce 2.70 g of ZnO, we need to use stoichiometry.
First, let's calculate the molar mass of ZnO:
M(Zn) + M(O) = 65.38 g/mol + 16.00 g/mol = 81.38 g/mol
Now, let's use the stoichiometry of the reaction to find the number of moles of ZnO:
2ZnO = 2 * 81.38 g/mol = 162.76 g/mol
Next, we need to find the number of moles of ZnO given the mass of 2.70 g:
n(ZnO) = mass / molar mass = 2.70 g / 81.38 g/mol = 0.0332 mol
Since the reaction is balanced, the mole ratio between oxygen gas (O2) and ZnO is 1:2. This means that 1 mole of oxygen gas reacts to produce 2 moles of ZnO.
So, the number of moles of O2 needed is:
n(O2) = 0.0332 mol * (1 mol O2 / 2 mol ZnO) = 0.0166 mol
Finally, we can find the mass of oxygen gas needed:
mass(O2) = n(O2) * molar mass(O2) = 0.0166 mol * 32.00 g/mol = 0.53 g
Therefore, 0.53 grams of oxygen gas must react to produce 2.70 grams of ZnO.