Magnesium reacts with hydrochloric acid according to the following equation: $Mg(s) + 2HCl(aq) -> MgCl_2(aq) +H_2(g)$. What mass of hydrogen will be obtained if 100 $cm^3$ of 2.00 mol $dm^-3$ HCl are added to 4.86 g of magnesium?

Md Ariful Islam

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Md Ariful Islam

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So you have the balanced equation, that's a good start.

So what we're looking for is the ratio of hydrogen gas produced per the limiting reactant, let's find that first, by getting this ugly DA out of the way.

I think you mean 2.00mol/dm^3, here, an annoying way to put (e.g. concentration) of that acidic solution.

$(2.00mol)/(dm^3)((10^-1 dm)/(cm))^3((cm^3)/(mL))((10^3 mL)/(L))$
$(2.00mol)/(dm^3)((10^-3 dm^3)/(cm^3))((cm^3)/(mL))((10^3 mL)/(L))$
$(2.00mol)/(L) = 2.00M$

What a waste of time, same value!

Now that we have the molarity, let's find the volume reacted.

$(100cm^3)/1((mL)/(cm^3)) = 100mL$

So, in a more convenient way, we have 100 mL of a 2.00M solution of $HCl$. Let's convert each of the reactants to moles, now:

HCl: $(2.00mol)/(.100L) = 20.0mol$

Mg(s): $4.86g ((mol)/(24.305g)) = 0.200mol$

Clearly, Mg is the limiting reactant , some quick math tells me all we need to fully react that molar amount is 0.400 mol of $HCl$.

$(0.200mol Mg)/1((H_2)/(Mg)) = 0.200mol H_2$

Then, the question asks for the mass, a matter of simply translating the moles of hydrogen gas to its mass in grams (or whatever else if the question specifies).

$(0.200mol)/1((2.016g)/(mol)) = 0.403g$

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