Sounds like an acid metal reaction to produce hydrogen gas, so I will assume you meant hydrochloric acid. Complete reaction indicates aluminum is limiting and drives the reaction.
2Al + 6HCl --> 2AlCl3 + 3H2
3.0 moles Al (3 moles H2/2 moles Al)
= 4.5 mole hydrogen gas produced
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4 Al + 3O2 ------> 2Al2O3 so 4 moles of Aluminium react with 6 moles of oxygen atoms or 3 moles of oxygen molecules. So 5 moles of Al reacts with 5/4 x 6 = 7.5 moles of oxygen atoms or 3.75 moles of oxygen molecules.
7.5 moles...
2,615 moles of aluminium oxide.
The answer is 1,25 moles iron(III) oxide.
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.
This is a mole stoichiometry problem. Start with the balanced equation for the synthesis of aluminum oxide: 4Al + 3O2 --> 2Al2O3. The ratio of aluminum to aluminum oxide in this equation is 4:2, or 2:1, so 5.23 moles Al means half that number for Al2O3, so about 2.62 moles of aluminum oxide will be produced.
1 mole Al(CN)3 = 105.03g _ moles = 163g 1 mole Al(CN)3 * 163g / 105.03g 163 moles/105.03 = 1.55 moles
The answer is 3,375 moles oxygen.
1.5 moles of O2 react with 2 moles of Al 2Al + 1.5 O2 = Al2O3.
Al+HCl===> AlCl3+H2 Is the reaction. You need &.2 moles of HCl.
2,615 moles of aluminium oxide.
The answer is 1,25 moles iron(III) oxide.
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.
Equation. 2Al + 3Cl2 -> 2AlCl3 one to one again 0.440 moles Al (2 moles AlCl3/2 moles Al) = 0.440 moles AlCl3 produced
Four:2 Al + 3 Cl2 --> 2 AlCl3so: 4 Al + 6 Cl2 --> 4 AlCl3
20 moles of Al will react with water to form 60 moles of H2 and 30 moles of O2, because of the stoichiometry 2:1 in getting H2 en O2 from water, and Al needing an oxidationstate of 3+, so 1 mole is oxidized by 3 moles of H2O.
10g Al(OH)3 is 0.12821 moles. The moles of hydroxide is thus, 0.38462 moles. If you have 1.5 moles per litre HCl and they react 1:1, you need 256.4ml
This reaction is:2 Al + 2 H2O + 2 NaOH = 2 NaAlO2 + 3 H2From 4 moles of Al 6 moles of hydrogen are obtained.
This is a mole stoichiometry problem. Start with the balanced equation for the synthesis of aluminum oxide: 4Al + 3O2 --> 2Al2O3. The ratio of aluminum to aluminum oxide in this equation is 4:2, or 2:1, so 5.23 moles Al means half that number for Al2O3, so about 2.62 moles of aluminum oxide will be produced.