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Question
- how many moles of carbon dioxide ($co_2$) will be produced from 3.019 moles of iron (iii) oxide, $fe_2o_3$?5. if $6.15 \times 10^{19}$ atoms of iron (fe) are formed, how many grams of carbon dioxide ($co_2$) are also produced? what is the correct conversion factor that belongs in the boxed area to solve the question?$\frac{6.15 \times 10^{19}\text{ atoms fe} \quad \quad \quad \frac{3\text{ mol }co_2}{4\text{ mol fe}} \quad \frac{44.01\text{ g }co_2}{1\text{ mol }co_2}}{}$ =(do not solve)
Step1 (Q4): Balance the chemical reaction
The reaction between iron(III) oxide and carbon monoxide (to produce iron and CO₂) is:
$$\ce{2Fe2O3 + 3C -> 4Fe + 3CO2}$$
Mole ratio: $\ce{2Fe2O3 : 3CO2}$
Step2 (Q4): Set up mole conversion
$$3.019\ \text{mol}\ \ce{Fe2O3} \times \frac{3\ \text{mol}\ \ce{CO2}}{2\ \text{mol}\ \ce{Fe2O3}}$$
Step3 (Q4): Calculate final moles
$$3.019 \times \frac{3}{2} = 4.5285\ \text{mol}$$
Step1 (Q5): Identify missing conversion factor
To convert Fe atoms to moles of Fe, use Avogadro's number: $\frac{1\ \text{mol}\ \ce{Fe}}{6.022 \times 10^{23}\ \text{atoms}\ \ce{Fe}}$
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- $4.529$ moles of $\ce{CO2}$ (rounded to 4 significant figures)
- The missing conversion factor is $\frac{1\ \text{mol}\ \ce{Fe}}{6.022 \times 10^{23}\ \text{atoms}\ \ce{Fe}}$