Chem Question ?

Question

The reaction of magnesium metal with HCl yields hydrogen gas and magnesium chloride. What is the volume in L of gas formed at 785 torr and 31 degrees Celsius from 1.42 g of Mg in excess HCl. ( First write the balanced equation )

Answer 1

CHESSLARUS made one mistake. According to the equation:

Mg + 2HCl --> MgCl2 + H2

one mol of Mg give 1 mol of H2. So 0.058 mol of Mg give 0.058 mol of H2, which occupies volume V:

V = n*R*T/p, V = 0.058*0.082*304/(785/760), V = 1.4 L approx.

Answer 2

1) First, the reaction that takes place here:

Mg + 2 HCl -----> MgCl2 + H2 (g)

and molecular mass of Magnesium:

M(Mg) = 24.3 g/mol

2) As we know the amount of Magnesium that react, we can calculate the number of moles of Mg:

n of Mg = 1.42 g / 24.3 g/mol = 0.0584 moles

3) Because the reaction took place with an excess of HCl we are sure that all Magnesium reacted stoichiometrically with the acid.
Thus, according to reaction above 1 mol of Magnesium reacts with 2 moles of HCl, then:

0.0584 moles of Mg reacted with 0.0584x2 moles of HCl and produced 0.0584 x 2 moles of H2. Hence, nH2 = 0.1168 moles

4) The volume can be easily computed using Gas Ideal Law:

PV = nRT
where: P = 785 torr
............T = 304.15 K
............R = 62.36 L torr / mol K..... (according to given units)
............n = 0.1168 moles

solving for V:

V = nRT/P = (0.1168)(62.36)(304.15)/785 torr

V = 2.822 L

That´s it!

Good luck!

NOTE: Dymos is right, I commited a mistake in the calculus above. Sorry for this lapsus. I think procedure is correct but I took two moles of H2 instead of ONLY ONE as the reaction states itself.

Right volume is

V = nRT/P = (0.0584)(62.36)(304.15)/785 torr
V = 1.411 L

Anyway, thanks for let me in!

Answer 3

This sounds like homework or a test, not what 1 usually needs 2 know. Look it up!! Ha Ha. I took 1 year of chemistry and don't want to take any more. Good luck on the test.