Question

A mixture of two metals found in Mendelejev's periodical table in different groups, reacted with 56cm³ of hydrogen on heating (measured at STP conditions) to produce two ionic compounds. These compounds were allowed to react with 270 mg of water but only one third of water reacted. A basic solution was formed in which the content of hydroxides was 30% by mass and at the same time deposited a precipitate with a mass that represented 59.05% of a total mass of the products formed by the reaction. After filtration the precipitate was heated and its mass decreased by 27 mg. When a stoichiometric amount of ammonium carbonate was added to the basic solution, a slightly soluble precipitate was obtained, at the same time ammonia was liberated and the content of hydroxides in the solution decreased to 16.81%. 

 Determine the metals in the starting mixture and their masses. 

Answer 1

Ionic hydrides are formed by combining of alkali metals or alkaline earth metals with hydrogen. In relation to the conditions in the task, there will be an alkali metal (M^I) as well as an alkaline earth metal (M^II) in the mixture. 

Equations

(1) M^I + 1/2 H₂ → M^IH 

(2) M^II + H₂ → M^IIH₂ 

(3) M^IH + H₂O → M^IOH + H₂ 

(4) M^IIH₂ + 2 H₂O → M^II(OH)₂ + 2H₂ 

reacted: 0.09 g H₂O, i. e. 0.005 mol unreacted: 0.18 g H₂O, i. e. 0.01 mol Since all hydroxides of alkali metals are readily soluble in water, the undissolved precipitate is M^II(OH)₂, however, it is slightly soluble in water, too. Thus, the mass of hydroxides dissolved in the solution: 

(5) m'(M^IOH + M^II(OH)₂) = Z 

Therefore: 

30 = Z/Z+0.18 x 100

Z = 0.077 g 

(6) m'(M^IOH + M^II(OH)₂) = 0.077 g 

It represents 40.95 % of the total mass of the hydroxides, i. e. the total mass of hydroxides is as follows: 

(7) m'(M^IOH + M^II(OH)2) = 0.077g x 100/40.95 = 0 188 g

The mass of solid M^II(OH)₂ : 

Precipitation with (NH₄CO₃): 

(11) Ca(OH)2 + (NH₄)₂CO₃ → CaCO₃ + 2 NH₃ + 2 H₂O According to (5) and (6) the mass of the solution was: 0.18 g + 0.077 g = 0.257 g After precipitation with (NH₄)₂CO₃ : 

16 .81 m(MIOH)/m(solution) x 100

Let us mark as n' the amount of substance of Ca(OH)₂ being present in the solution. M(Ca(OH)₂) = 74 g mol^-1 

 Taking into account the condition in the task as well as equation (11), we get: 

The total amount of substance of Ca(OH)₂ (both in the precipitate and in the solution): 

(12) (Ca(OH) ) = 0.111 g/74 g mol-1 + 5 x10-4  mol 0.002 mol (i. e. 0.148 g)

According to equations (3) and (4): 

n(H₂O) = 0.004 mol (for M^IIH₂) 

n(H₂O) = 0.001 mol (for M^IH) 

n(M^IOH) = 0.001 mol 

According to equations (7) and (11): 

Composition of the mixture: 0.002 mol Ca + 0.001 mol Na 

or 0.080 g Ca + 0.023 g Na