What volume of hydrogen gas is produced when 39.34 g of magnesium metal reacts with 2890 mL of phosphoric acid at STP? *

V = 36.22 L

Explanation:

In this case, we need to write the overall reaction which is the following:

3Mg(s) + 2H₃PO₄(l) > Mg₃(PO₄)₂(aq) + 3H₂(g)

Now, if we have the volume of acid, we need to get the density of the same to transform this volume to mass. In this case, the density of phosphoric acid is 1.88 g/mL

the mass of the acid is:

d = m/V ---> m = d * V

m = 2890 * 1.88 = 5443.2 g

Now, in order to solve this, we need to do stochiometry and to do that, we need to use the moles and then get the limiting reactant and the excess so we can calculate the volume of hydrogen. To get the moles we need the molar masses.

MM of Mg = 24.31 g/mol

MM of H₃PO₄ = 97.994 g/mol

Now we can calculate the moles:

moles Mg = 39.34 / 24.31 = 1.618 moles

moles H₃PO₄  = 5443.2 / 97.994 = 55.546 moles

Now according to the overall reaction:

3Mg(s) + 2H₃PO₄(l) > Mg₃(PO₄)₂(aq) + 3H₂(g)

According to this, 3 moles of Mg reacts with only 2 moles of acid, but we have 55.546 moles of acid and 1.618 of magnesium. Clearly the limiting reactang is the magnesium, and we can prove it like this:

3 moles Mg > 2 moles acid

1.618 > X acid

X = 1.618 * 2 / 3 = 1.079 moles of acid

As you can see we have way more moles of acid, so the limiting reactant is the magnesium.

moles of magnesium have the same mole ratio than the moles of hydrogen, so we can assume the moles of magnesium would be the same moles produced of hydrogen:

moles Mg  = moles H₂ = 1.618 moles

Now, as we are working at STP (T = 273 K; P = 1 atm), we can use the ideal gas equation to solve for the volume:

PV = nRT

V = nRT/P

We have the moles, the pressure, and volume. And R  is gas constant which is 0.082 L atm / mol K. Replacing these data the volume of hydrogen would be:

V = 1.618 * 0.082 * 273 / 1

V = 36.22 L

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