Question #a7d7b

2 Answers
Feb 8, 2018

9.29 g Li must be added to liquid water to obtain 15.0 L H2 Gas.

Explanation:

At STP, one mol gas occupies 22.4 L. Therefore, 15L/22.4L gives us .6696 mol H2 gas produced. The stoichiometric mole ratio of Li reactants to H2 products is 2:1. Finally, there are 6.94 g Li per mole of Li. Setting this all up, we get:
(1 mol H2)/(22.4 L H2) x (15 L H2)/1= .6696 mol H2
(.6696 mol H2) x (2 mol Li)/(1 mol H2) x (6.94 g Li)/(1 mol Li) =9.29 g Li.

All our units cancel out nicely and we get our answer.

Feb 8, 2018

You will need 9.30 g of Li. Details of how to solve the problem are below.

Explanation:

Here's what you do:

First convert the given amount into moles

15.0 L -:22.4 L/"mol" = 0.67 mol

Next, check the equation to find the mole ratio between the substance you know (H_2) and the substance you must determine (Li). THis is given by the coefficients of the balanced equation:

2 moles of Li produce 1 mole of H_2.

Now set up two equivalent ratios. One contains the unknown and known substances, the other has the coefficients:

("moles of Li")/("moles of" H_2)=2/1

Solve for the unknown:

"moles of Li"=2/1xx("moles of" H_2)

"moles of Li"=2xx0.67 = 1.34 moles

Finally, convert this answer into grams (or whatever unit you are asked to give):

1.34 "mol" xx6.94 g/"mol" = 9.30 g