Apure organic amide product (2.85 g) which has a molecular weight of 285.34 g/mol was fully dissolved in 20ml of the boiling solvent, hexanes. this organic amide product then underwent a recrystallization when the solvent was cooled to 0° c. the recrystallized organic product (2.62 grams) was obtained after vacuum filtration. what is the solubility of the organic amide product in the solvent, hexanes, at 0° c?

The solubility of the organic amide product in the solvent hexane is 0.0403 moles/L at 0° C

Explanation:

When 2.85gm dissolved in 20ml of boiling solvent

After re crystallization 2.62gm was obtained

so, that the dissolved amount at 0° C is

2.85 - 2.62

= 0.23 gm in 20ml hexane

Mole of amide

Solubility

               = 0.0403 moles/L

Therefore, the solubility of the organic amide product in the solvent hexane is 0.0403 moles/L at 0° C

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2