27.7k views
3 votes
onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency, which reactant is in excess and how much is leftover? The molar mass of CaCN2 is 80.11 g/mol. The molar mass of CaCO3 is 100.09 g/mol.

User Urkle
by
7.8k points

1 Answer

6 votes

Answer : The excess reactant is,
H_2O

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of
CaCN_2 = 105.0 g

Mass of
H_2O = 78.0 g

Molar mass of
CaCN_2 = 80.11 g/mole

Molar mass of
H_2O = 18 g/mole

Molar mass of
CaCO_3 = 100.09 g/mole

First we have to calculate the moles of
CaCN_2 and
H_2O.


\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=(105.0g)/(80.11g/mole)=1.31moles


\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=(78.0g)/(18g/mole)=4.33moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,


CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3

From the balanced reaction we conclude that

As, 1 mole of
CaCN_2 react with 3 mole of
H_2O

So, 1.31 moles of
CaCN_2 react with
1.31* 3=3.93 moles of
H_2O

From this we conclude that,
H_2O is an excess reagent because the given moles are greater than the required moles and
CaCN_2 is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.


\text{ Mass of excess reactant}=\text{ Moles of excess reactant}* \text{ Molar mass of excess reactant}(H_2O)


\text{ Mass of excess reactant}=(0.4moles)* (18g/mole)=7.2g

Thus, the leftover amount of excess reagent is, 7.2 grams.

User Steve Sanbeg
by
8.0k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.