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Consider the following reaction:
Mg(s) + 2HCl(aq)
->
MgCl2 (aq) +H2(8)

If 0.475 g Mg reacts with 124.95 mL of 1.08 M HCl, how many mol of H2 gas will be produced?

Calculate the initial moles of Mg and HCl and find the limiting reactant?


Sagot :

Torrisjossiidn

Answer:The balanced equation of Mg(s) + 2HCl(aq) ==> MgCl2(aq) + H2(g) does NOT tell you how many moles of Mg or H2 are present.  It only tells you the MOLE RATIO of the reactants and products.  Thus, 1 mole Mg reacts with 2 moles HCl to produce 1 mole of hydrogen gas.

Explanation:The balanced equation of Mg(s) + 2HCl(aq) ==> MgCl2(aq) + H2(g) does NOT tell you how many moles of Mg or H2 are present.  It only tells you the MOLE RATIO of the reactants and products.  Thus, 1 mole Mg reacts with 2 moles HCl to produce 1 mole of hydrogen gas.

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