This value is 38,9 mol.
2 moles of aluminum to 3 moles of oxygen
4 moles
33 grams aluminum (1 mole Al/26.98 grams) = 1.2 moles of aluminum ================
No moles of oxygen are produced by complete combustion of propane. Oxygen is CONSUMED, not produced. For combustion of 4 moles of propane, it will use 20 moles of oxygen.
1,075 moles of aluminium chloride are obtained.
2 moles of aluminum to 3 moles of oxygen
4 moles
4 moles
33 grams aluminum (1 mole Al/26.98 grams) = 1.2 moles of aluminum ================
No moles of oxygen are produced by complete combustion of propane. Oxygen is CONSUMED, not produced. For combustion of 4 moles of propane, it will use 20 moles of oxygen.
1,075 moles of aluminium chloride are obtained.
depends on how much aluminum oxide you have, 1gram, 2 billion Kg? how many? cant find the number of moles of oxygen without knowing the mass of the al203
there are two moles produced in potassium nitrate.
12 moles KClO3 (3 moles O/1 mole KClO3) = 36 moles of oxygen.
Equation. 2Al + 3Cl2 -> 2AlCl3 one to one again 0.440 moles Al (2 moles AlCl3/2 moles Al) = 0.440 moles AlCl3 produced
6 moles
The balanced chemical equation for the reaction between aluminum (Al) and oxygen (Oβ) to form aluminum oxide (AlβOβ) is: [ 4 \text{ Al} + 3 \text{ O}_2 \rightarrow 2 \text{ Al}_2\text{O}_3 ] According to the balanced equation, 4 moles of aluminum (Al) produce 2 moles of aluminum oxide (AlβOβ). Therefore, if 4.0 moles of aluminum completely react, it will produce ( \frac{2}{4} \times 4.0 ) moles of aluminum oxide. Calculate that to find the answer.