Question

how much cuo would form if 12.9 grams of cu reacts with 3.2 grams of o2?

reaction	reactant(s)	product(s)
mass	12.9 g	3.2 g	?

Explanation:

Step1: Write the balanced chemical equation

$2Cu + O_2
ightarrow 2CuO$

Step2: Calculate the molar masses

The molar mass of $Cu$ is $M_{Cu}=63.55\ g/mol$, the molar mass of $O_2$ is $M_{O_2} = 32\ g/mol$, and the molar mass of $CuO$ is $M_{CuO}=63.55 + 16=79.55\ g/mol$.

Step3: Calculate the number of moles of reactants

The number of moles of $Cu$, $n_{Cu}=\frac{m_{Cu}}{M_{Cu}}=\frac{12.9\ g}{63.55\ g/mol}\approx0.203\ mol$.
The number of moles of $O_2$, $n_{O_2}=\frac{m_{O_2}}{M_{O_2}}=\frac{3.2\ g}{32\ g/mol}= 0.1\ mol$.

Step4: Determine the limiting reactant

From the balanced - equation, the mole ratio of $Cu$ to $O_2$ is $2:1$. For $0.1\ mol$ of $O_2$, the amount of $Cu$ required is $n_{Cu\ required}=2\times n_{O_2}=2\times0.1\ mol = 0.2\ mol$. Since we have $0.203\ mol$ of $Cu$, $O_2$ is the limiting reactant.

Step5: Calculate the mass of $CuO$ formed

From the balanced - equation, the mole ratio of $O_2$ to $CuO$ is $1:2$. So the number of moles of $CuO$ formed, $n_{CuO}=2\times n_{O_2}=2\times0.1\ mol = 0.2\ mol$.
The mass of $CuO$ formed, $m_{CuO}=n_{CuO}\times M_{CuO}=0.2\ mol\times79.55\ g/mol = 15.91\ g$.

Answer:

$15.91\ g$