Question

in an experiment conducted by a student, 333 g h₂ reacts with 300.03 g o₂ to produce h₂o.
2 h₂ + 1 o₂ → 2 h₂o
mole ratio: 2:1
yield = actual yield / theo × 100%

Explanation:

Step1: Calculate moles of reactants

First, find moles of $H_2$ and $O_2$. Moles of $H_2=\frac{333g}{2.016g/mol}$ (molar - mass of $H_2$ is $2.016g/mol$), moles of $O_2=\frac{300.03g}{32.00g/mol}$ (molar - mass of $O_2$ is $32.00g/mol$).

Step2: Determine limiting reactant

From the balanced equation $2H_2 + O_2
ightarrow2H_2O$, the mole - ratio of $H_2$ to $O_2$ is $2:1$. Calculate the amount of $H_2O$ produced from each reactant. For $H_2$, moles of $H_2O$ produced $=\text{moles of }H_2\times1$ (from mole - ratio). For $O_2$, moles of $H_2O$ produced $=\text{moles of }O_2\times2$. The reactant that produces the least amount of $H_2O$ is the limiting reactant.

Step3: Calculate theoretical yield

Based on the limiting reactant, calculate the theoretical yield of $H_2O$ in grams. Multiply the moles of $H_2O$ (from limiting reactant) by the molar - mass of $H_2O$ ($18.02g/mol$).

Step4: Calculate percent yield

Percent yield $=\frac{\text{actual yield}}{\text{theoretical yield}}\times100\%$. Since the actual yield is not given in the problem, we can only set up the formula for percent yield as $\text{Percent yield}=\frac{\text{actual yield}}{\text{theoretical yield}}\times100\%$, where theoretical yield is calculated from the limiting reactant.

Answer:

The problem is incomplete as the actual yield is not provided. But the steps to calculate the percent yield are as above. First, find moles of reactants, then the limiting reactant, then the theoretical yield, and finally use the percent - yield formula.