3.62P - Chemistry: Molecular Nature Of Matter And Change

Question

Reactants D and E form product F. Write a detailed Plan to find the mass of F when 27 g of D reacts with 31 g of E.

Step-by-Step Solution

Answer

By comparing the molar ratio of reactants present in that of needed in the balanced chemical equation, now denoting the coefficient of the limiting reactant as l_r. Using the number of moles of the limiting reactant we have got, we proceed as follows:

$\mathrm{Mass} \mathrm{of} \mathrm{F}=\mathrm{mol} \mathrm{LR}\times \frac{\mathrm{fmolF}}{1_{\mathrm{\Gamma }}\mathrm{molLR}}\times \frac{{\mathrm{M}}_{\mathrm{w}}\mathrm{ofF}\left(\mathrm{g}\right)\mathrm{F}}{1\mathrm{molF}}=\mathrm{gF}$

Step1. Finding the limiting reactant.

The chemical equation according to the question can be written as :

$\mathrm{dD}+\mathrm{eE}\to \mathrm{fF}$

Now the given mass of reactants D and E are converted by multiplying the reciprocal of their molar masses. By this process we will identify that which reactant is the limiting reactant(LR), By comparing the molar ratio of reactants present in that of needed in the balanced chemical equation, now denoting the coefficient of the limiting reactant as l_r. Using the number of moles of the limiting reactant we have got, we proceed as follows:

$\mathrm{Mass} \mathrm{of} \mathrm{F}=\mathrm{mol} \mathrm{LR}\times \frac{\mathrm{fmolF}}{{\mathrm{I}}_{\mathrm{\Gamma }}\mathrm{molLR}}\times \frac{{\mathrm{M}}_{\mathrm{w}}\mathrm{ofF}\left(\mathrm{g}\right)\mathrm{F}}{1\mathrm{molF}}=\mathrm{gF}$

Step2. Final result

The chemical equation according to the question can be written as :

$\mathrm{dD}+\mathrm{eE}\to \mathrm{fF}$

Now the given mass of reactants D and E are converted by multiplying the reciprocal of their molar masses. By this process we will identify that which reactant is the limiting reactant(LR), By comparing the molar ratio of reactants present in that of needed in the balanced chemical equation, now denoting the coefficient of the limiting reactant as l_r. Using the number of moles of the limiting reactant we have got, we proceed as follows:

$\mathrm{Mass} \mathrm{of} \mathrm{F}=\mathrm{mol} \mathrm{LR}\times \frac{\mathrm{fmolF}}{{\mathrm{I}}_{\mathrm{\Gamma }}\mathrm{mol} \mathrm{LR}}\times \frac{{\mathrm{M}}_{\mathrm{w}} \mathrm{of} \mathrm{F}\left(\mathrm{g}\right)\mathrm{F}}{1\mathrm{molF}}=\mathrm{gF}$