Welcome to Westonci.ca, where finding answers to your questions is made simple by our community of experts. Discover reliable solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
Sagot :
Answer: 73.4 g of [tex]Fe_2O_3[/tex] are formed when 51.3 grams of iron react completely with oxygen.
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} Fe=\frac{51.3g}{56g/mol}=0.92moles[/tex]
The balanced chemical reaction is:
[tex]4Fe+3O_2\rightarrow 2Fe_2O_3[/tex]
As [tex]Fe[/tex] is the limiting reagent as it limits the formation of product.
According to stoichiometry :
4 moles of [tex]Fe[/tex] produce = 2 moles of [tex]Fe_2O_3[/tex]
Thus 0.92 moles of [tex]Fe[/tex] will produce=[tex]\frac{2}{4}\times 0.92=0.46moles[/tex] of [tex]Fe_2O_3[/tex]
Mass of [tex]Fe_2O_3=moles\times {\text {Molar mass}}=0.46moles\times 159.69g/mol=73.4g[/tex]
Thus 73.4 g of [tex]Fe_2O_3[/tex] are formed when 51.3 grams of iron react completely with oxygen.
Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.
