Ask Question
9 July, 12:18

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.

+2
Answers (1)
  1. 9 July, 12:23
    0
    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.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “3) Lithium metal (Li) react with hydrosulfuric acid (HS) to produce hydrogen gas and magnesium chloride (Li 2 S). How many grams of Lithium ...” in 📗 Chemistry if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers