I answered this for you yesterday. You can go back and find my answer. Here are the instructions although my post yesterday was different than this (but the same answer).
Convert 0.45 mol Al to mols C needed. Use the coefficients in the balanced equation for this.
Convert mols C needed to atoms C remembering that 1 mol C will contain 6.02E23 atoms C.
Aluminum reacts with oxygen gas to produce aluminum oxide. How many moles of aluminum oxide would you expect to produce if you reacted 14.5 moles of oxygen gas with excess aluminum? 4Al +3O2 --> 2Al2O3 14.5 mol O2 x 4 mol Al /
Given the following: 4Al+ 3O2---> 2Al2O3 Which reactant is limiting if 0.32 mol Al and 0.26 mol O2 are available? I got 0.24 mol Al and 0.35 mol O2 when I did my work. Therefore Al is limiting. Can you tell me if I am right.
Given 2Al2O3 (s) --> 4Al(s) + 3O2 (g) (standard enthalpy change= 3351.4 kJ) a) What is the heat of formation of aluminum oxide? How do I find heat of formation from standard enthalpy change? I know how to do it when I'm given
Aluminum metal is used in many products. It quickly reacts with oxygen in the reaction 4Al (s) + 3O2 (g)--> 2Al2O3 (s). Which component is different on each side of the equation arrow? a. mass b. atoms c. moles d. all are the
Given ∆Hºrxn = -1670 kJ/mol for 2Al(s) + (3/2)O2(g) --> Al2O3(s), determine ∆Hº for the reaction 2Al2O3(s) --> 4Al(s) + 3O2(g). I flipped Al(s) + (3/2)O2(g) --> Al2O3(s) to Al2O3(s) --> A(g) + (3/2)O2(s) and