Question
Asked Feb 19, 2020
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2 Al (s) + 2 KOH (aq) + 6 HOH (l) ---> 2 KAl(OH)4 (aq) + 3 H2 (g)

Under experimental procedures #2 and #3, the experiment calls for reacting aluminum with water and potassium hydroxide. See above for the equation. A student using quantities:

3.00 grams Al

15.0 mL of 8.00 M KOH

75.0 mL distilled HOH

Which of the reactants is the limiting reactant? (Mathematically document this.) Show your work and set up.

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Expert Answer

Step 1

Given:

3.00 grams Al

15.0 mL of 8.00 M KOH

75.0 mL distilled H2O

 

Chemical reaction is given by,

2 Al (s) + 2 KOH (aq) + 6 HOH (l) ---> 2 KAl(OH)4 (aq) + 3 H2 (g)

Step 2

Calculation for moles of Al:

Chemistry homework question answer, step 2, image 1

Calculation for moles of KOH:

Chemistry homework question answer, step 2, image 2

Calculation for moles of H2O:

Mass = density × volume

Mass = 1 g/mL × 75.0 mL

Mass of H2O = 75.0 g

Chemistry homework question answer, step 2, image 3

...

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