8) A solution, made by dissolving 0.80 g of an unknown nitrogen stimulant in 12.0 grams of water, boils at 100.21 °C at 760 mm Hg. Given the molal boiling point elevation constant of water as Kb =
0.512 °C/m, the best value below for the molar mass (MM) of the nitrogen compound is:
A. 88.0 g/mole
B. 162 g/mole
C. 246 g/mole
D. 197 g/mole E) 375 g/mole

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

there are two significant figures.

the right answer is c i just did the question.