To determine how many grams of aluminum (Al) are required to consume 17.5 grams of sodium hydroxide (NaOH), we need to follow these steps:
1. Determine the molar masses of the reactants:
- Molar mass of NaOH: Sodium (Na) has a molar mass of 23 g/mol, oxygen (O) has a molar mass of 16 g/mol, and hydrogen (H) has a molar mass of 1 g/mol.
[tex]\[
\text{Molar mass of NaOH} = 23 + 16 + 1 = 40 \, \text{g/mol}
\][/tex]
- Molar mass of Al: Aluminum (Al) has a molar mass of 27 g/mol.
2. Calculate the number of moles of NaOH in 17.5 grams:
[tex]\[
\text{Moles of NaOH} = \frac{\text{Mass of NaOH}}{\text{Molar mass of NaOH}} = \frac{17.5 \, \text{g}}{40 \, \text{g/mol}} = 0.4375 \, \text{moles}
\][/tex]
3. Use the balanced chemical equation to find the mole ratio between NaOH and Al:
The balanced chemical equation is:
[tex]\[
6 \text{NaOH} (aq) + 2 \text{Al} (s) \longrightarrow 3 \text{H}_2 (g) + 2 \text{Na}_3 \text{AlO}_3 (aq)
\][/tex]
From the balanced equation, the mole ratio of NaOH to Al is:
[tex]\[
\frac{6 \text{moles NaOH}}{2 \text{moles Al}} = \frac{6}{2} = 3
\][/tex]
Which means 3 moles of NaOH react with 1 mole of Al.
4. Calculate the number of moles of Al needed:
Since 3 moles of NaOH react with 1 mole of Al:
[tex]\[
\text{Moles of Al needed} = \frac{1 \, \text{mole Al}}{3 \, \text{moles NaOH}} \times 0.4375 \, \text{moles NaOH} = 0.145833 \, \text{moles Al}
\][/tex]
5. Calculate the mass of Al needed:
[tex]\[
\text{Mass of Al} = \text{Moles of Al} \times \text{Molar mass of Al} = 0.145833 \, \text{moles} \times 27 \, \text{g/mol} = 3.9375 \, \text{g}
\][/tex]
So, the mass of aluminum needed to consume 17.5 grams of sodium hydroxide is approximately 3.9375 grams.
Hence, the final result is:
[tex]\[
\boxed{3.9375 \, \text{grams}}
\][/tex]
