IDNLearn.com: Your trusted platform for finding reliable answers. Our experts are available to provide accurate, comprehensive answers to help you make informed decisions about any topic or issue you encounter.
Sagot :
well, let’s use stoichiometry to find out.
30.01g Fe2O3 x (1molFe203/159.69gFeNo3)
* (4molFe/2molFeNO3) x (55.845gFe/1molFe) = 20.99 g Fe needed to produce 30.01g of Fe2O3
So, this is true!
Confused where I got the values from?
- the 30.01g Fe2O3 comes from the problem.
- the 159.69g Fe2NO3 is the molar mass found by added the mass of each element and how much of each element is present.
-the 4mol of Fe and the 2mol Fe2NO3 are found in the balanced equation.
-the 55.845 g of Fe is the molar mass of fe on the periodic table.
hope this helps!
30.01g Fe2O3 x (1molFe203/159.69gFeNo3)
* (4molFe/2molFeNO3) x (55.845gFe/1molFe) = 20.99 g Fe needed to produce 30.01g of Fe2O3
So, this is true!
Confused where I got the values from?
- the 30.01g Fe2O3 comes from the problem.
- the 159.69g Fe2NO3 is the molar mass found by added the mass of each element and how much of each element is present.
-the 4mol of Fe and the 2mol Fe2NO3 are found in the balanced equation.
-the 55.845 g of Fe is the molar mass of fe on the periodic table.
hope this helps!
Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. Accurate answers are just a click away at IDNLearn.com. Thanks for stopping by, and come back for more reliable solutions.