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 : The excess reactant is,

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of = 105.0 g

Mass of = 78.0 g

Molar mass of = 80.11 g/mole

Molar mass of = 18 g/mole

Molar mass of = 100.09 g/mole

First we have to calculate the moles of and .

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

From the balanced reaction we conclude that

As, 1 mole of react with 3 mole of

So, 1.31 moles of react with moles of

From this we conclude that, is an excess reagent because the given moles are greater than the required moles and 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.

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

? you expect us to answer this? ? this is your schoolwork, i don't think we can with this, i think it would considering being

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.