Question

molybdenum(iv) oxide can be prepared by reaction of molybdenum(vi) oxide with hydrogen at high temperature: moo₃ + h₂ → moo₂ + h₂o. the reaction of 402.7 g of moo₃ with excess h₂ yields 316.2 g of moo₂. calculate the theoretical yield of moo₂ (assuming complete reaction) and its percentage yield. theoretical yield = blank g. percentage yield = blank %. 3 item attempts remaining. use the references to access important values if needed for this question.

Explanation:

Step1: Calculate molar masses

The molar mass of $MoO_3$: $M_{MoO_3}=95.96 + 3\times16.00=143.96\ g/mol$. The molar mass of $MoO_2$: $M_{MoO_2}=95.96+2\times16.00 = 127.96\ g/mol$.

Step2: Determine moles of $MoO_3$

The moles of $MoO_3$, $n_{MoO_3}=\frac{m_{MoO_3}}{M_{MoO_3}}=\frac{402.7\ g}{143.96\ g/mol}\approx2.797\ mol$.

Step3: Calculate theoretical yield of $MoO_2$

From the balanced - chemical equation $MoO_3 + H_2
ightarrow MoO_2+H_2O$, the mole ratio of $MoO_3$ to $MoO_2$ is 1:1. So the moles of $MoO_2$ produced theoretically is $n_{MoO_2}=n_{MoO_3}\approx2.797\ mol$. The theoretical mass of $MoO_2$, $m_{theo}=n_{MoO_2}\times M_{MoO_2}=2.797\ mol\times127.96\ g/mol\approx357.1\ g$.

Step4: Calculate percentage yield

The percentage yield formula is $\text{Percentage yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times100\%$. Given actual yield $m_{actual} = 316.2\ g$. So $\text{Percentage yield}=\frac{316.2\ g}{357.1\ g}\times100\%\approx88.6\%$.

Answer:

Theoretical yield = 357.1 g
Percentage yield = 88.6%