Your can dissolve an aluminum soft drink can in an(aq) base such as potassium hydroxide. 2Al(s)+2KOH(aq)+6H2O→2KAl(OH)4(aq)+3H2(g) If you place 2.05 g of alumimum in a beaker with 185 mL of 1.35 M KOH, will any aluminum remain(show all calculations). What mass of KAl(OH)4 is produced?

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Your can dissolve an aluminum soft drink can in an(aq) base such as potassium hydroxide.  2Al(s)+2KOH(aq)+6H2O→2KAl(OH)4(aq)+3H2(g)  If you place 2.05 g of alumimum in a beaker with 185 mL of 1.35 M KOH, will any aluminum remain(show all calculations). What mass of KAl(OH)4 is produced?

Aubree Mcintyre · Accepted Answer

mole Al=(2.05g)(1mole26.98g)=7.60×10−2 mole Al mole KOH = (0.185 L)(1.35 mole/L) = 0.250 mole KOH According to the equation, for every mole of KOH you need one moleof Al. Al is therefore the limiting reactant and no aluminum will remain after the reaction. As Al is the limiting reactant, calculate KAl(OH)4 as follows: mole KAl(OH)4=(7.60×10−2 mole Al)(2 mole KAl(OH)42 mole Al)=7.60×10−2mole KAl(OH)4 mass KAl(OH)4=(7.60×10−2)(134.07gmole)=10.2gKAl(OH)4

Your can dissolve an aluminum soft drink can in an(aq) base such as potassium hydroxide. 2Al(s)+2KOH(aq)+6H_2O\to2KAl(OH)_4(aq)+3H_2(g) If you place 2.05 g of alumimum in a beaker with 185 mL of 1.35 M KOH, will any aluminum remain(show all calculations). What mass of KAl(OH)_4 is produced?

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