Question

Calculate the weight of FeO formed from 2g of VO and 5.75 g of Fe₂O₃. Also report the limiting reagent. (Atomic mass V = 51.4, O = 16, Fe = 55.9 g) $2VO+3Fe_{2}O_{3}\longrightarrow 6FeO+V_{2}O_{5}$

Answer 1

To determine:  

a) The weight of FeO which is obtained when 2g of VO is mixed with 5.75 g of ${ Fe }_{ 2 }{ O }_{ 3 }$

b) The limiting reagent involved in the formation of FeO

Given Data: The balanced equation of the chemical reaction involved in the formation of FeO from VO and ${ Fe }_{ 2 }{ O }_{ 3 }$:

                   $2VO\quad +\quad 3Fe_{ 2 }O_{ 3 }\quad \longrightarrow \quad 6FeO\quad +\quad V_{ 2 }O_{ 5 }$

Weight of VO: 2g

Weight of ${ Fe }_{ 2 }{ O }_{ 3 }$: 5.75 g

Atomic mass of V = 51.4 g

Atomic mass of O = 16 g

Atomic mass of Fe = 55.9 g

Calculation:

Step 1: From the balanced equation, we can find the number of moles of each reactant and product involved in the chemical reaction. Consequently, the relative weights of each reactant and product is as follows:

$2VO\quad \quad \quad +\quad \quad 3Fe_{ 2 }O_{ 3 }\quad \longrightarrow \quad 6FeO+V_{ 2 }O_{ 5 }$

$(2\times 67)\quad \quad \quad \quad (3\times 160)\quad \quad \quad \quad \quad \quad (6\times 72)$

$134g\qquad \qquad 480g\qquad \qquad \quad \quad \quad 432g$

Step 2: Let us randomly choose any of the reactants as limiting reagent, for example, let us say VO

Applying unitary method, we get the mass of ${ Fe }_{ 2 }{ O }_{ 3 }$ which is required in reacting with 2 g of VO, as follows

          $\frac { 2\quad \times \quad 480 }{ 134 } \quad =\quad 7.164\quad g$

Step 3:  The required amount of mass of ${ Fe }_{ 2 }{ O }_{ 3 }$ ( i.e., 7.64 g) is greater than the mass of ${ Fe }_{ 2 }{ O }_{ 3 }$ given in question (i.e., 5.75g ). Hence the 5.75 g will be completely used up in the reaction and the reaction will come to an end.  Thus, it can be concluded that ${ Fe }_{ 2 }{ O }_{ 3 }$ is the limiting reagent.

Step 4: Here since the entire quantity of ${ Fe }_{ 2 }{ O }_{ 3 }$ gets used up in the formation of FeO, we can determine the weight of FeO using unitary method as follows:

480 g of ${ Fe }_{ 2 }{ O }_{ 3 }$ is used for forming 432 g of FeO (as shown in Step 1)

So, if 5.75 g of ${ Fe }_{ 2 }{ O }_{ 3 }$ is used, $\frac { 5.75\quad \times \quad 432 }{ 480 } \quad =\quad 5.18\quad g$ of FeO is formed