Question

Consider the reaction of potassium carbonate with calcium nitrate to form potassium nitrate and calcium carbonate: Suppose 50ml of a 0.250M potassium carbonate solution was mixed with 100ml of a 0.175M calcium nitrate solution. What is the maximum amount of calcium carbonate that could be obtained?

Answer 1

maximum amount of calcium carbonate=12.5 mili mole

Step-by-step explanation:

First write the balance chemical reaction:

`K_2CO_3 +Ca(NO_3)_2\rightarrow2KNO_3+CaCO_3`

quantity of
`K_2CO_3`: 50ml and 0.25M

quantity of
`Ca(NO_3)_2`: 100ml and 0.175M

`mili\, mole\, =volume\, in\, ml * molarity`

`mili\, mole\, of\, K_2CO_3 =50 * 0.25=12.5`

`mili\, mole\, of\, Ca(NO_3)_2 =100 * 0.175=17.5`

From the above balance equation it is clearly that:

1 mole of potassium carbonate react completely with 1 mole calcium nitrate

hence

12.5 mili mole of potassium carbonate react completely with 12.5 mili mole calcium nitrate but we have 17.5 mili mole of calcium nitrate therefore calcium nitrate will be the excess reactant and potassium carbonate will be the limiting reactant and quantiy of product depends on the limiting reactant i.e. potassium carbonate .

From the above balance equation:

1 mole of potassium carbonate gives 1 mole calcium carbonate

hence

12.5 mili mole will give 12.5 mili mole of calcium carbonate

maximum amount of calcium carbonate=12.5 mili mole