Question

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.

Answer 1

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.