Question

to produce zinc metal, carbon monoxide reacts with zinc (ii) oxide according to: zno + co → zn + co2. the carbon monoxide, however, is initially prepared from: 2c + o2 → 2co. how many grams of zinc metal (zn) may be prepared if 66.20g of c is used with excess amount of zno? 721.6g 1443g 360.8g 6.073g numeric 10 points. how many moles of lithium are there in 11 moles of lithium oxide? answer

Explanation:

Step1: Calculate moles of C

The molar - mass of C is $M_{C}=12.01g/mol$. The number of moles of C, $n_{C}=\frac{m_{C}}{M_{C}}$, where $m_{C} = 66.20g$. So, $n_{C}=\frac{66.20g}{12.01g/mol}=5.512mol$.

Step2: Determine moles of CO from the C - reaction

From the reaction $2C + O_{2}
ightarrow2CO$, the mole ratio of C to CO is 1:1. So, the number of moles of CO produced, $n_{CO}=n_{C}=5.512mol$.

Step3: Determine moles of Zn from the ZnO - CO reaction

From the reaction $ZnO + CO
ightarrow Zn+CO_{2}$, the mole ratio of CO to Zn is 1:1. So, the number of moles of Zn, $n_{Zn}=n_{CO}=5.512mol$.

Step4: Calculate mass of Zn

The molar - mass of Zn is $M_{Zn}=65.38g/mol$. The mass of Zn, $m_{Zn}=n_{Zn}\times M_{Zn}=5.512mol\times65.38g/mol = 360.8g$.

For the second part:

Step1: Determine the formula of lithium oxide

The formula of lithium oxide is $Li_{2}O$.

Step2: Calculate moles of Li in lithium oxide

In 1 mole of $Li_{2}O$, there are 2 moles of Li. So, in 11 moles of $Li_{2}O$, the number of moles of Li is $n_{Li}=2\times11mol = 22mol$.

Answer:

For the first question: 360.8g
For the second question: 22mol