Question

or the given reaction, if we react 4.9 grams of KOH and 12.8 grams of H3PO4 , which is the limiting reactant?3 KOH + H3PO4 → K3PO4 + 3 H2OSelect one:a.K3PO4b.H3PO4c.H2Od.KOH

Answer 1

`KOH`

Step-by-step explanation:

Here, we want to get the limiting reactant

The limiting reactant is the reactant that produces less amount of the product

Firstly, let us get the number of moles of the reactants

To get that, we have to divide the mass of the reactant by the molar mass

For Potassium Hydroxide, we have its molar mass as 56 g/mol

Thus, the number of moles of KOH that reacted will be:

`(4.9)/(56)\text{ = 0.0875 mol}`

From the equation of reaction:

3 moles KOH produced 1 mole K3PO4

0.0875 KOH will produce:

`(0.0875*1)/(3)\text{ = 0.0292 mol}`

For H2O, 0.0875 mole will be produced as the two have equal coefficients

For H3PO4, its molar mass is 98 g/mol

The number of moles that reacted will be:

`(12.8)/(98)\text{ = 0.131 mol}`

K3PO4 and H3PO4 have the same coefficients. Thus, the number of moles of K3PO4 produced will be 0.131 mol

For H2O, we have:

`3*0.131\text{ = 0.393 mol}`

What was produced with KOH was lesser and that makes it the limiting reactant