Question

In a aqueous solution 0.5 mole of `"CaClO"_2"` is reacted with 0.5 mole of sodium phosphorus. The number of moles of calcium phosphorus formed is?

Answer 1

0.167 `mol` will be the number of moles for Calcium phosphide formed.

Explanation:

The balanced equation of the reaction above is:
`3Ca(ClO)_2` + `2Na_3P` `rarr` `6NaCl` + `Ca_3P_2` + `3O_2`

Even though both of the reactants are 0.5 mole, but the coefficient of both reactants are different after the equation is being balanced. Hence, we can calculate `n` of `Ca` and `n` of `P`.

`n` of `Ca` = `0.5 mol` / `3` = `0.167mol`
`n` of `P` = `0.5mol` / `2` = `0.25mol`

From the calculation, we can see that sodium phosphide, `Na_3P` is in excess, and calcium hypochlorite, `Ca(ClO)_2` is the limiting factor.

Therefore, the number of mole of calcium phosphide, `Ca_3P_2` will be `0.167mol`.

Answer 2

A maximum of `"0.2 mol Ca"_3("PO"_4)_2"` can be produced. The limiting reactant is calcium hypochlorite `("CaClO"_2")`.

Explanation:

There is no sodium phosphorus nor calcium phosphorus. There are sodium phosphide and sodium phosphate, as well as calcium phosphide and calcium phosphate. Based on their properties, I'm going to go with sodium phosphate and calcium phosphate.

Balanced equation

`"3CaClO"_2("aq") + "2Na"_3"PO"_4("aq")` `rarr` `"3Na"_2"ClO"_2("aq") + "Ca"_3"(PO"_4)_2("s")darr"`

The down arrow means that the solid is a precipitate.

Mol calcium phosphate from 0.5 mol calcium hypochlorite.

To calculate mol calcium phosphate produced from given mol `"CaClO"_2"`, multiply the given mol `"CaClO"_2"` by the mol ratio between `"CaClO"_2"` and `"Ca"_3"(PO"_4)_2"` from the balanced equation, with `"Ca"_3"(PO"_4)_2"` in the numerator.

`0.5color(red)cancel(color(black)("mol CaClO"_2))xx(1"mol Ca"_3("PO"_4)_2)/(3color(red)cancel(color(black)("mol CaClO"_2)))="0.2 mol Ca"_3("PO"_4)_2` (rounded to one significant figure)

Mol calcium phosphate produced from 0.5 mol sodium phosphate.

To calculate mol calcium phosphate produced from the given mol `"Na"_3"PO"_4`, multiply the given mol `"Na"_3"PO"_4` by the mol ratio between `"Ca"_3("PO"_4)_2"`, with `"Ca"_3("PO"_4)_2"` in the numerator.

`0.5color(red)cancel(color(black)("mol Na"_3"PO"_4))xx(1"mol Ca"_3("PO"_4)_2)/(2color(red)cancel(color(black)("mol Na"_3"PO"_4)))="0.3 mol Ca"_3("PO"_4)_2` (rounded to one significant figure)

This is a limiting reactant question. The reactant that produces the least moles of the product calcium phosphate is the limiting reactant. In this case, the limiting reactant is calcium hypochlorite `("CaClO"_2")`.