Question

5.         From the information below, determine the mass of substance C that is formed if 45.0 g of substance A reacts with 23.0 g of substance B, assuming that the reaction between A and B goes to completion.

A. Substance A is a solid that consists of a Group 13 (3A) metal (boron is not a metal!) and carbon; the m% C in substance A is 57.2%. It reacts with substance B to form substances C and D. 4.0 x 10¹⁶ formula units of substance A weighs 8.371 micrograms.

B. 47.9 g of substance B contains 5.36 g H and 42.5 g O.

C.   When 10.0 g of substance C is burned in excess oxygen, 33.8 g of CO₂and 6.92 g of H₂O are produced. The molar mass of substance C is M_C = ~26 g/mol.

D. Substance D is the hydroxide of the metal in substance A.

Answer 1

a) 4.0 x 10¹⁶ formula units of substance A weighs 8.371 micrograms.

So Avogadra no of formula units make molar mass /formula mass

a) The molecular mass of A   = ( ( 8.371x10^-6 ) x 6.02x10²³ )/4x10¹⁶

                                             = 126 gram per formula unit approx

The compound on hydrolysis contains acetylene ion so the formula will be Al₂(C₂)₃

The percentage of C is 57.14% in the compund and the metal of 13 group is Al

B ) 47.9 g of substance B contains 5.36 g H and 42.5 g O.

The compound contains       H       O        Only we can determine formula as

i) divide by atomic mass of each           5.36/1              42.5/16 =2.6625

ii) Divide each by smallest                     approx 2              1

So its H₂O = B

c) When 10.0 g of substance C is burned in excess oxygen, 33.8 g of CO₂ and 6.92 g of H₂O are produced. The molar mass of substance C is M_C = ~26 g/mol.

It should be acetylene = C

HCCH   as per mass given above 26.

D) The hydroxide is of aluminium so its Al(OH)₃

              SO we Have following reaction

-----------------------------------------------------------------

The reaction of A & B will be as follows :

Applying stiochiometric calculations we can find the mass of each

Al₂(C₂)₃      + 6 H₂O -------> 2 Al(OH)3       +    3 Acetylene

.35 mole        1.27 mole

water is the limiting reagent so all calculation according to amount of water.

Al(OH)3 moles = 1.27/3          moles of acetylene    = 1.27/2