1700 kJ are released when 2 moles Fe2O3 are formed. So 1700/2 = 850 kJ are released for 1 mole Fe2O3 formed. How many moles Fe2O3 do you have from 453 kJ? That will be 453/850 = ??
Convert that to grams. '
Check my thinking.
Use the following equation to answer the question...
4Fe(s)+3O2(g)---> 2Fe2O3(g)
Heat = -1700kJ
How many grams of rust form when 453kJ of energy is released?
Convert that to grams. '
Check my thinking.
Here are the steps to solve this problem:
1. Calculate the moles of energy released:
- Divide the given heat (-1700 kJ) by the molar enthalpy (heat) of the reaction. In this case, the molar enthalpy is -1700 kJ since it is given in the equation.
- So, moles of energy released = -1700 kJ / -1700 kJ/mol = 1 mole of energy released.
2. Determine the stoichiometry of the reaction:
- From the balanced equation, we can see that 4 moles of iron (Fe) react with 3 moles of oxygen (O2) to form 2 moles of rust (Fe2O3).
- This means that 2 moles of rust are formed per 4 moles of iron.
- Simplifying, we can say that 1 mole of rust is formed per 2 moles of iron.
3. Calculate the moles of rust formed:
- Since we know that 1 mole of energy released corresponds to the formation of 1 mole of rust, the moles of rust formed is also 1 mole.
4. Determine the molar mass of rust:
- Fe2O3 consists of two iron atoms (2 x atomic mass of iron) and three oxygen atoms (3 x atomic mass of oxygen).
- The atomic mass of iron (Fe) is 55.845 g/mol, and the atomic mass of oxygen (O) is 15.999 g/mol.
- Therefore, the molar mass of Fe2O3 is (2 x 55.845 g/mol) + (3 x 15.999 g/mol) = 159.687 g/mol.
5. Convert moles of rust to grams:
- Multiply the moles of rust formed (1 mole) by the molar mass of Fe2O3 (159.687 g/mol):
- Mass of rust formed = 1 mole x 159.687 g/mol = 159.687 g.
So, when 453 kJ of energy is released, 159.687 grams of rust will form according to the given equation.