Question

1. According to the following reaction, how many grams of iron(III) oxide are required for the complete reaction of 28.0 grams of hydrochloric acid?

hydrochloric acid (aq) + iron(III) oxide (s) water (l) + iron(III) chloride (aq)


2. According to the following reaction, how many grams of hydrobromic acid are needed to form 23.4 grams of bromine?

hydrobromic acid (aq) hydrogen (g) + bromine (l)

Answer 1

1) First we should equalize the equation:

6HCl + Fe₂O₃ = 2FeCl₃ + 3H₂O

Now, we should found the amount in moles of HCl and that of Fe₂O₃ would be 6 times less:

$n(HCl)=\frac{28.0g}{36.5g/mol}=0.767mol$

$n(Fe_2O_3)=\frac{1}{6}*0.767mol=0.128mol$

Thus, the mass of Fe₂O₃ is equal to:

$m(Fe_2O_3)=160g/mol*0.128mol=20.5g$

2) Lets equalize it:

2HBr = H₂ + Br₂

Now, lets calculate the amount of bromine in moles and the amount of HBr would be 2 times Br₂ amount:

$n(Br_2)=\frac{23.4g}{160g/mol}=0.146mol$

$n(HBr)=2*0.146mol=0.292mol$

Now, it is the right time to calculate the mass of HBr:

$m(HBr)=81g/mol*0.292mol=23.7g$

Thus, the mass of HBr required is equal to 23.7g.