Sagot :
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 [tex]\text{2 mol NaOH} \bumpeq \text{2 mol} \: \text{NaAlO}_{2}[/tex], the number of moles of NaAlO₂ produced is
[tex]\text{moles of} \: \text{NaAlO}_{2} = \text{126.4 g NaOH} × \frac{\text{1 mol NaOH}}{\text{39.998 g NaOH}} × \frac{\text{2 mol} \: \text{NaAlO}_{2}}{\text{2 mol NaOH}}[/tex]
[tex]\text{moles of} \: \text{NaAlO}_{2} = \text{3.160158 mol}[/tex]
[tex]\boxed{\text{moles of} \: \text{NaAlO}_{2} = \text{3.160 mol}}[/tex]
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).
[tex]\text{mass of} \: \text{NaAlO}_{2} = \text{3.160158 mol} \: \text{NaAlO}_{2} × \frac{\text{81.97 g} \: \text{NaAlO}_{2}}{\text{1 mol} \: \text{NaAlO}_{2}}[/tex]
[tex]\boxed{\text{mass of} \: \text{NaAlO}_{2} = \text{259.0 g}}[/tex]
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 [tex]\text{2 mol Al} \bumpeq \text{2 mol} \: \text{NaAlO}_{2}[/tex], the number of moles of NaAlO₂ produced is
[tex]\text{moles of} \: \text{NaAlO}_{2} = \text{97.70 g Al} × \frac{\text{1 mol Al}}{\text{26.98 g Al}} × \frac{\text{2 mol} \: \text{NaAlO}_{2}}{\text{2 mol Al}}[/tex]
[tex]\text{moles of} \: \text{NaAlO}_{2} = \text{3.621 mol}[/tex]
Step 3: Determine the limiting reagent.
Since NaOH produced less number of moles of NaAlO₂
[tex]\boxed{\text{NaOH is the limiting reagent.}}[/tex]
Solution: (d)
Step 1: Calculate the mass of excess reagent consumed.
The excess reagent in this reaction is Al.
[tex]\text{mass of Al consumed} = \text{3.160158 mol} \: \text{NaAlO}_{2} × \frac{\text{2 mol Al}}{\text{2 mol} \: \text{NaAlO}_{2}} × \frac{\text{26.98 g Al}}{\text{1 mol Al}}[/tex]
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
[tex]\boxed{\text{mass of Al left = 12.44 g}}[/tex]
[tex]\\[/tex]
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