Answer to Question #333380 in General Chemistry for mei

Question

an aqueous solution of NaOH can dissolve an aluminum sheet in the reaction:

2 Al + 2 NaOH + 2H2O → 2 NaAlO₂ + 3H₂

Answer 1

Given:

mass of NaOH = 126.4 g

mass of Al = 97.70 g

balanced chemical equation: 2 Al + 2 NaOH + 2H₂O → 2 NaAlO₂ + 3 H₂

Required:

moles of NaAlO₂

mass of NaAlO₂

limiting reagent

mass of excess reagent left

Solution: (a)

Step 1: Calculate the molar mass of NaOH.

molar mass of NaOH = (22.99 g/mol × 1) + (16.00 g/mol × 1) + (1.008 g/mol × 1)

molar mass of NaOH = 39.998 g/mol

Step 2: Calculate the number of moles of NaAlO₂ produced by using dimensional analysis (factor-label method).

Since , the number of moles of NaAlO₂ produced is

Solution: (b)

Step 1: Calculate the molar mass of NaAlO₂.

molar mass of NaAlO₂ = (22.99 g/mol × 1) + (26.98 g/mol × 1) + (16.00 g/mol × 2)

molar mass of NaAlO₂ = 81.97 g/mol

Step 2: Calculate the mass of NaAlO₂ produced by using dimensional analysis (factor-label method).

Solution: (c)

Step 1: Calculate the molar mass of Al.

molar mass of Al = 26.98 g/mol × 1

molar mass of Al = 26.98 g/mol

Step 2: Calculate the number of moles of NaAlO₂ produced by using dimensional analysis (factor-label method).

Since , the number of moles of NaAlO₂ produced is

Step 3: Determine the limiting reagent.

Since NaOH produced less number of moles of NaAlO₂

Solution: (d)

Step 1: Calculate the mass of excess reagent consumed.

The excess reagent in this reaction is Al.

mass of Al consumed = 85.26 g

Step 2: Calculate the mass of excess reagent left.

mass of Al left = initial mass of Al - mass of Al consumed

mass of Al left = 97.70 g - 85.26 g