Question

1.Calculate how many moles of NH3 form when each quantity of reactant completely reacts according to the equation:

3N2H4(l)→4NH3(g)+N2(g)

4.10 mol N2H4

65.8 g N2H4

Hydrobromic acid dissolves solid iron according to the following reaction:

2.Fe(s)+2HBr(aq)→FeBr2(aq)+H2(g)

What mass of HBr (in g) would you need to dissolve a 3.2 −g pure iron bar on a padlock?

Express your answer using two significant figures.

What mass of H2 would be produced by the complete reaction of the iron bar?

Express your answer using two significant figures.

Answer 1

1)

a)

from reaction,

moles of NH3 formed = (4/3)*moles of N2H4 reacted

= (4/3)*4.10 moles

= 5.47 mol

Answer: 5.47 mol

b)

Molar mass of N2H4,

MM = 2*MM(N) + 4*MM(H)

= 2*14.01 + 4*1.008

= 32.052 g/mol

mass(N2H4)= 65.8 g

number of mol of N2H4,

n = mass of N2H4/molar mass of N2H4

=(65.8 g)/(32.052 g/mol)

= 2.053 mol

from reaction,

moles of NH3 formed = (4/3)*moles of N2H4 reacted

= (4/3)* 2.053 moles

= 3.34 mol

Answer: 3.34 mol

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