Question #99eac

Start by taking a look at the balanced chemical equation for this single replacement reaction

#"Mg"_text((s]) + 2"HCl"_text((aq]) -> "MgCl"_text(2(aq]) + "H"_text(2(g]) uarr#

Now, chemical equations make use of stoichiometric coefficients to express the mole ratios that exist between the reactants and the products when a reaction takes place.

In essence, the coefficients written before the chemical symbols of the species that take part in a reaction tell you how many moles of each are involved in the equation.

It's important to remember that a coefficient of #1# is never written in the balanced chemical equation! In your case, you would have

#"Mg" <=> 1 xx "Mg" -># one mole of magnesium

#2"HCl" <=> 2 xx "HCl" -># two moles of hydrochloric acid

#"MgCl"_2 <=> 1 xx "MgCl"_2 -># one mole of magnesium chloride

#"H"_2 <=> 1 xx "H"_2 -># one mole of hydrogen gas

So, the reaction needs two moles of hydrochloric acid for every one mole of magnesium metal. Likewise, the reaction produces one mole of magnesium chloride and one mole of hydrogen gas for every one mole of magnesium metal that takes part in the reaction.

So, if you react one mole of magnesium metal, you will get one mole of hydrogen gas, hence the #1:1# mole ratio that exists between the two chemical species.

To get the amount of gas in grams, you can use hydrogen gas' molar mass.

#1 color(red)(cancel(color(black)("mole H"_2))) * "2.016 g"/(1color(red)(cancel(color(black)("mole H"_2)))) ~~ "2 g H"_2#

To get the volume of gas produced, you need to know the conditions for pressure and temperature.

At STP, Standard Temperature and Pressure, one mole of any ideal gas occupies exactly #"22.7 L"#, so this much hydrogen gas would occupy #"22.7 L"# at #0^@"C"# and #"100 kPa"#.