Question

Question #a7d7b

Answer 1

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.

Answer 2

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`