Consider the following reaction: li3n (s) + 3h2o (l) → nh3 (g) + 3lioh (l) if you need to make 24 g lioh, how many grams of li3n must you react with excess water?

11.63 g

Explanation:

The reaction given is balacend:

Li₃N(s) + 3H₂O(l) → NH₃(g) + 3LiOH(l)

The molar masses of the interest compounds are:

Li₃N = 34.83 g/mol

LiOH = 23.95 g/mol

By the stoichiometry of the reaction:

1 mol of Li₃N 3 moles of LiOH

Transforming these values to mass multiplying the number of moles by the molar mass:

34.83 g of Li₃N 71.85 g of LiOH

x                         24 g of LiOH

By a simple direct three rule:

71.85x = 835.92

x = 11.63 g of Li₃N

i dont know the answe

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole