Question

A sample of aluminum metal is placed in a graduated cylinder. It is noted that 5.50 mL of water is displaced by the aluminum. The aluminum is then reacted with excess nitric acid to produce aluminum nitrate and hydrogen gas. Given the density for aluminum is 2.702 g/mL, how many grams of aluminum nitrate are produced in the reaction? 2Al(s) + 6HNO3(aq) + 2Al(NO3)3(aq) + 3H2(g)

Answer 1

volume of Al = volume of water displaced

= 5.50 mL

Mass of Al reacted = density * volume

= 2.702 g/mL * 5.50 mL

= 14.861 g

Molar mass of Al = 26.98 g/mol

mass of Al = 14.86 g

mol of Al = (mass)/(molar mass)

= 14.86/26.98

= 0.5508 mol

According to balanced equation

mol of Al(NO3)3 formed = moles of Al

= 0.5508 mol

Molar mass of Al(NO3)3,

MM = 1*MM(Al) + 3*MM(N) + 9*MM(O)

= 1*26.98 + 3*14.01 + 9*16.0

= 213.01 g/mol

mass of Al(NO3)3 = number of mol * molar mass

= 0.5508*2.13*10^2

= 1.173*10^2 g

Answer: 117 g