1. Aluminum reacts with oxygen to produce aluminum oxide which can be used as an adsorbent, desiccant or catalyst for organic reactions.

4Al(s) + 3O2(g) → 2Al2O3(s)

A mixture of 82.49 g of aluminum ( Picture = 26.98 g/mol) and 117.65 g of oxygen ( Picture = 32.00 g/mol) is allowed to react. Identify the limiting reactant and determine the mass of the excess reactant present in the vessel when the reaction is complete.

2. Sodium hydroxide, also known as caustic soda, is used to neutralize acids and to treat cellulose in making of cellophane. Calculate the number of moles of solute in 1.875 L of 1.356 M NaOH solution.

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  1. mols Al = g/atomic mass = approx 3.06 but you should do this more accurately (as well as all of the other calculations below. I estimate.)
    Convert to mols O2 using the coefficients in the balanced equation. That's 3.05 x (3 mols O2/4 mol Al) = 2.29
    Do you have that much O2? You have 117.65/32 = 3.7 mols; therefore, there is enough to react with all of the Al and that makes Al the limiting reagent.

    How much O2 is left. You reacted with 2.29 mols O2. You had 3.7 mol initially. mols left = 3.7-2.29 = ? and that convert to grams is mass O2 remaining. Check my work.

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  2. Consider 2H2 + O2 �¨ H2O. To produce 0.895 g water, how many grams of O2 are required?

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  3. gross

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