the aluminum will dissolve. Generating lots of heat during the reaction
Al+HCl===> AlCl3+H2 Is the reaction. You need &.2 moles of HCl.
2 Al + 6 HCl = 2AlCl3 + 3H2Aluminum Chloride salt with hydrogen gas emitted.
To find the molarity, first calculate the moles of HCl by using the molar mass of Al to convert grams of Al to moles, since HCl and Al react in a 1:1 ratio. Then, divide the moles of HCl by the volume in liters to get the molarity. Molarity = moles of solute / volume of solution in liters.
The balanced chemical equation for the reaction of aluminum with hydrochloric acid (HCl) to form hydrogen gas (H2) and aluminum chloride (AlCl3) is: 2 Al + 6 HCl → 2 AlCl3 + 3 H2
2Na + 2H2O = 2NaOH + H2 2Na + 2HCl = 2NaCl + H2 Na+H2O= NaOH2
When aluminum is added to dilute hydrochloric acid (HCl), a chemical reaction occurs. The aluminum reacts with the HCl to form aluminum chloride and hydrogen gas. The reaction is exothermic and produces bubbles of hydrogen gas as it proceeds.
The mass of hydrochloric acid needed to react with 87.7 grams of aluminum can be calculated using stoichiometry. The balanced chemical equation for the reaction between hydrochloric acid (HCl) and aluminum (Al) is 2Al + 6HCl → 2AlCl3 + 3H2. By applying stoichiometry, you'll find that the molar mass ratio between Al and HCl is 1:6. Therefore, the amount of HCl needed to react with 87.7 grams of Al is: (87.7 grams Al) x (6 moles HCl / 1 mole Al) x (36.46 g HCl / 1 mole HCl) = 151.63 grams of HCl.
To determine the moles of HCl that can be neutralized by 5.63 grams of Al(OH)3, you first need to calculate the molar mass of Al(OH)3. Then, using the balanced chemical equation for the neutralization reaction between Al(OH)3 and HCl, you can determine the mole ratio between Al(OH)3 and HCl. Finally, you can use this ratio to calculate the moles of HCl that can be neutralized by 5.63 grams of Al(OH)3.
Balanced equation first. 2Al + 6HCl >> 2AlCl3 + 3H2 (find limiting reactant ) 5 mole Al (6 mole HCl/2mole Al) = 15 mole HCl 5 mole HCl (2 mole Al/6 mole HCl) = 1.7 mole Al ( Al looks short of what is required, so will drive reaction ) 5 moles Al (3 moles H2/2 moles Al) = 7.5 moles of H2 produced.
balanced equation: 2 Al + 6 HCl → 2 AlCl3 + 3 H2 we know the total amount of hcl is 87.7grams now we want to find out how many grams one part hcl is: so 87.7grams of hcl, divided by 6 = 14.61g so then the amount of al needed can be worked out by multiplying 14.61g by 2. (al) we get 29.22grams of al This is not correct. The balancing can't be done in grams it needs to be in moles, so the molecular mass of Al and HCl needs to be looked up and used. See the link to Stoichiometry.
The reaction between aluminum (Al) and hydrochloric acid (HCl) is a single replacement reaction. The aluminum metal replaces the hydrogen in hydrochloric acid, producing aluminum chloride and hydrogen gas.
Simple methods:- reaction of Zn with HCl- reaction of Al with NaOH