Al donates 3 electrons per mole to be oxidised.
H2SO4 needs (accepts) 2 electrons per mole when oxidising reactive metallics like Aluminum.
(in the following: units are in brackets)
So 2.5 mole Al reacts with 2.5(molAl) * 3(el's/molAl) / 2(el's/molH2SO4) =
7.5(el's) / 2(el's/molH2SO4) = 3.75 mole H2SO4
24 mols of aluminum must have reacted since aluminum oxide is Al2O3
1 mole sulfuric acid for 1 mole calcium chloride
To calculate the mass of aluminum (Al) needed to completely react with 135 g of iron, you first need to determine the molar ratio between them from the balanced chemical equation. The balanced equation for the reaction between aluminum and iron is 2Al + 3Fe2O3 -> 3Fe + Al2O3. From this ratio, you can calculate that 2 moles of aluminum (Al) react with 3 moles of iron (Fe). Next, calculate the molar mass of aluminum (Al) and use it to convert the given mass of iron to moles. Finally, use the molar ratio to find the mass of aluminum needed.
These reagents doesn't react.
0,5 moles Cl-
24 mols of aluminum must have reacted since aluminum oxide is Al2O3
1 mole sulfuric acid for 1 mole calcium chloride
To calculate the mass of aluminum (Al) needed to completely react with 135 g of iron, you first need to determine the molar ratio between them from the balanced chemical equation. The balanced equation for the reaction between aluminum and iron is 2Al + 3Fe2O3 -> 3Fe + Al2O3. From this ratio, you can calculate that 2 moles of aluminum (Al) react with 3 moles of iron (Fe). Next, calculate the molar mass of aluminum (Al) and use it to convert the given mass of iron to moles. Finally, use the molar ratio to find the mass of aluminum needed.
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.
When aluminum reacts completely with oxygen to form aluminum oxide, the molar ratio is 4:2. This means that 4 moles of Al produce 2 moles of Al2O3. So, if 5.23 moles of Al are reacted, half as many moles of Al2O3 can be made, which is 2.615 moles.
These reagents doesn't react.
0,5 moles Cl-
Since the ratio of moles of Al to moles of Al2O3 is 4:2, if 5.23 mol Al completely reacts, 2.615 mol Al2O3 can be made.
When 4 moles of aluminum react with an excess of chlorine gas, 4 moles of aluminum chloride are produced. This is because the balanced chemical equation for the reaction is: 2Al + 3Cl2 -> 2AlCl3 This means that 2 moles of aluminum react with 3 moles of chlorine gas to produce 2 moles of aluminum chloride, so 4 moles of aluminum will produce 4 moles of aluminum chloride.
The mole ratio of aluminum to oxygen in aluminum oxide (Al2O3) is 4:3, which means for every 4 moles of aluminum, there are 3 moles of oxygen.
0.8 moles Explanation: from the equation we can see, 2 mole A l is needed to react completely with 3 mole F e O so, 3 moles of F e O needs 2 moles A l so, 1 mole F e O needs 2 3 moles A l so, 1.2 mol F e O needs 2 Γ 1.2 3 moles A l = 0.8 moles A l
The balanced equation for the reaction is: 3H2 + N2 -> 2NH3 From the balanced equation, we can see that 3 moles of hydrogen are needed to react completely with 1 mole of nitrogen. So if there are 3 moles of nitrogen, you would need 9 moles of hydrogen to react completely.