Question

If 155 grams of potassium (K) reacts with 122 grams of potassium nitrate (KNO3), what is the limiting reagent? 2KNO3 + 10K 6K2O + N2

Answer 1

K = 39 g/mol
KNO3 = 101 g/mol

2 KNO3 + 10 K = 6 K2O + N2

2 x 101 g KNO3 ---------- 10 x 39 g K
122 g KNO3 -------------- ??

122 x 10 x 39 / 2 x 101 =

47580 / 202 => 235.54 g  of K

 ( KNO3 is Excess reagent ) 

2 x 101 g KNO3 ---------- 10 x 39 g K
??  --------------------------- 155 g K

155 x 2 x 101 / 10 x 39 =

31310 / 390 => 80.28 g of KNO3   ( K is limiting reagent )

hope this helps!

Answer 2

Answer : The limiting reagent is, potassium

Explanation : Given,

Mass of potassium = 155 g

Mass of potassium nitrate = 122 g

Molar mass of potassium = 39 g/mole

Molar mass of potassium nitrate = 101 g/mole

First we have to calculate the moles of $K$ and $KNO_3$.

$\text{Moles of }K=\frac{\text{Mass of }K}{\text{Molar mass of }K}=\frac{155g}{39g/mole}=3.97moles$

$\text{Moles of }KNO_3=\frac{\text{Mass of }KNO_3}{\text{Molar mass of }KNO_3}=\frac{122g}{101g/mole}=1.21moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$2KNO_3+10K\rightarrow 6K_2O+N_2$

From the balanced reaction we conclude that

As, 2 moles of $KNO_3$ react with 10 mole of $K$

So, 1.21 moles of $KNO_3$ react with $\frac{10}{2}\times 1.21=6.05$ moles of $K$

That means, in the given balanced reaction, $K$ is a limiting reagent because it limits the formation of products and $KNO_3$ is an excess reagent.

Hence, the potassium is the limiting reagent.