3) Lithium metal (Li) react with hydrosulfuric acid (HS) to produce hydrogen gas and magnesium chloride (Li 2 S). How many grams of Lithium metal are required to react completely with excess acid to produce 4.5 L of Hydrogen at 315K and 1.258 atm? Remember to balance the equation.

3.066g

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2Li + H2S - > Li2S + H2

Step 2:

Data obtained from the question.

Volume (V) of H2 = 4.5L

Temperature (T) = 315K

Pressure (P) = 1.258 atm

Note:

Gas constant (R) = 0.0821atm. L/Kmol

Number of mole (n) of H2 = ... ?

Step 3:

Determination of the number of mole of H2 produced.

This can be obtained by using the ideal gas equation as follow

PV = nRT

Divide both side by RT

n = PV / RT

n = 1.258 x 4.5 / 0.0821 x 315

n = 0.219 mole

Therefore, 0.219 mole of H2 is produced.

Step 4:

Determination of the number of mole of Li that will produce 0.219 mole of H2.

This is shown below:

2Li + H2S - > Li2S + H2

From the balanced equation above,

2 moles of Li reacted to produce 1 mole of H2.

Therefore, Xmol of Li will react to produce 0.219 mole of H2 i. e

Xmol of Li = 2 x 0.219

Xmol of Li = 0.438 mole

Step 5:

Conversion of 0.438 mole of Li to grams.

Number of mole of Li = 0.438 mole

Molar Mass of Li = 7g/mol

Mass = number of mole x molar Mass

Mass of Li = 0.438 x 7

Mass of Li = 3.066g

Therefore, 3.066g if Li is needed for the reaction.