Question

Shift 855 #. Consider the reaction 2 KNO3 + 10K - 6K2O + N2. If 155 grams of k reacted with 122 grams of KNO3, what is the limiting reagent? KNO3 = 101.11 g/mol, K = 39.10 g/mol a) KNO3 b)K c) neither d) not enough information 21.Fe.O. - 2Fe + A1 Os. Suppose that 0,450 moles of Fe2O, are

Answer 1

Molar mass of KNO3 = 101.11 g/mol

mass(KNO3)= 122.0 g

use:

number of mol of KNO3,

n = mass of KNO3/molar mass of KNO3

=(1.22*10^2 g)/(1.011*10^2 g/mol)

= 1.207 mol

Molar mass of K = 39.1 g/mol

mass(K)= 155.0 g

use:

number of mol of K,

n = mass of K/molar mass of K

=(1.55*10^2 g)/(39.1 g/mol)

= 3.964 mol

Balanced chemical equation is:

2 KNO3 + 10 K ---> 6 K2O + N2

2 mol of KNO3 reacts with 10 mol of K

for 1.207 mol of KNO3, 6.033 mol of K is required

But we have 3.964 mol of K

so, K is limiting reagent

Answer: b