Aluminum reacts with chlorine gas to form aluminum chlonde via the following rea

Question

Aluminum reacts with chlorine gas to form aluminum chlonde via the following reaction: 2Al(s) + 3C12 (g) rightarrow 2AlCl3 (s) you are given 14.0 g of aluminum and 19 0 g of chlorine gas If you had excess chlorine, how many moles of of aluminum chloride could be produced from 14.0 g of aluminum? Express your answer to three significant figures and include the appropriate units.If you had excess aluminum, how many moles of aluminum chlondo could be produced from 19.0 g of chlorine gas. Cl2? Express your answer to three significant figures and include tho appropriate units.

Explanation / Answer

part A )

mass of Al = 14 g

molar mass of Al = 27 g/mol

moles of Al = 14 / 27 = 0.5185

2Al + 3 Cl2 ---------------> 2AlCl3

2 mol Al gives ---------------> 2 mol AlCl3

0.5185 mol Al gives ------------> 2 x 0.5185 /2 = 0.5185 mol

AlCl3 moles = 0.518

part B)

Cl2 moles = 19 / 71 = 0.2676

2Al + 3 Cl2 ---------------> 2AlCl3

3 mol Cl2 gives -----------------> 2 mol AlCl3

0.2676 mol Cl2 gives ------------> x mol AlCl3

x = 2 x 0.2676 / 3

x = 0.178 mol

AlCl3 moles = 0.178

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