Question

How many moles of magnesium chloride are produced when 4.39 moles of magnesium are placed in a beaker

containing only 250.0 grams of hydrochloric acid?

Answer 1

`n_(MgCl_2)=3.43mol MgCl_2`

Step-by-step explanation:

Hello!

In this case, since the reaction between magnesium and hydrochloric acid is:

`Mg+2HCl\rightarrow MgCl_2+H_2`

Whereas there is a 1:1 and 1:2 mole ratio between magnesium and magnesium chloride and hydrochloric acid and magnesium chloride respectively; it is possible to compute the yielded moles of product by each reactant and then determine which moles are correct based on the limiting reactant; therefore, we proceed as follows:

`n_(MgCl_2)^(by\ Mg)=4.39molMg*(1molMgCl_2)/(1molMg) =4.39molMgCl_2\\\\n_(MgCl_2)^(by\ HCl)=250.0gHCl*(1molHCl)/(36.46gHCl)*(1molMgCl_2)/(2molHCl)=3.43mol MgCl_2`

Thus, since the hydrochloric acid yields the fewest moles of product, we infer it is the limiting reactant, so the correct number moles of magnesium chloride are 3.43 mol.

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