Answer : The limiting reactant is [tex]H_3PO_4[/tex].
Explanation : Given,
Mass of [tex]Cr[/tex] = 0.50 g
Mass of [tex]H_3PO_4[/tex] = 0.75 g
Molar mass of [tex]Cr[/tex] = 52 g/mole
Molar mass of [tex]H_3PO_4[/tex] = 98 g/mole
First we have to calculate the moles of [tex]Cr[/tex] and [tex]H_3PO_4[/tex].
[tex]\text{Moles of }Cr=\frac{\text{Mass of }Cr}{\text{Molar mass of }Cr}=\frac{0.50g}{52g/mole}=0.00962moles[/tex]
[tex]\text{Moles of }H_3PO_4=\frac{\text{Mass of }H_3PO_4}{\text{Molar mass of }H_3PO_4}=\frac{0.75g}{98g/mole}=0.00765moles[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
[tex]2Cr+2H_3PO_4\rightarrow 2CrPO_4+3H_2[/tex]
From the balanced reaction we conclude that
As, 2 moles of [tex]H_3PO_4[/tex] react with 2 mole of [tex]Cr[/tex]
So, 0.00765 moles of [tex]H_3PO_4[/tex] react with [tex]\frac{2}{2}\times 0.00765=0.00765[/tex] moles of [tex]Cr[/tex]
From this we conclude that, [tex]Cr[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]H_3PO_4[/tex] is a limiting reagent and it limits the formation of product.
Hence, the limiting reactant is [tex]H_3PO_4[/tex].
