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

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!

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BarrettArcher BarrettArcher · Answer 2 · 2019-05-13T02:14:15-04:00

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.