We will firstly convert the mass into moles to dfeine the relationship using mole ratio.
![Fe_2O_(3(s))+3C_((s))\rightarrow2Fe_((s))+3CO_((g))](https://img.qammunity.org/2023/formulas/chemistry/college/b862b5pixg1j8p5a8aucw53g8cwlxznws6.png)
![\begin{gathered} _nC_((s))=\frac{mass}{molar\text{ }mass} \\ \\ _nC_((s))=\frac{33.0\text{ }g}{12\text{ }gmol^(-1)} \\ _nC_((s))=2.75\text{ }mol \end{gathered}](https://img.qammunity.org/2023/formulas/chemistry/college/funlzeuit05iq5ofcan6ryhee2apldw62y.png)
The mole ratio between C and CO is 3:3 or 1:1. Therefore 2.75 mol of carbon would produce 2.75 mol of carbon monoxide. Using this relationship we will convert moles to mass.
![\begin{gathered} mass\text{ }CO=moles* molar\text{ }mass \\ mass\text{ }CO=2.75\text{ }moles*28\text{ }gmol^(-1) \\ mass\text{ }CO=77g \end{gathered}](https://img.qammunity.org/2023/formulas/chemistry/college/q8xxycss3tjuprj0sw8ez8fjrurw9i2mix.png)
Answer: 77.0g of CO are produced when 33.0g of C reacts.