# Chapter 3 - Chemical Reactions - Questions for Review and Thought - Topical Questions - Page 149e: 69c

A total of 1.67g of $Al$ was not consumed.

#### Work Step by Step

1. Calculate the number of moles of $Cl_2$: 35.45* 2 = 70.9g/mol $4.05g \times \frac{1 mol}{ 70.9g} = 0.0571mol (Cl_2)$ - The balanced reaction is: $2Al + 3Cl_2 -- \gt Al_2Cl_6$ The ratio of $Cl_2$ to $Al$ is 3 to 2: $0.0571 mol (Cl_2) \times \frac{ 2 mol(Al)}{ 3 mol (Cl_2)} = 0.0381mol (Al)$ 2. Calculate the mass of $Al$: 26.98* 1 = 26.98g/mol $0.0381 mol \times \frac{ 26.98 g}{ 1 mol} = 1.03g (Al)$ 3. The initial $Al$ mass was 2.70g, and 1.03g was consumed: 2.70g - 1.03g = 1.67g

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