Question

Instructions:Select the correct answer. 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

2KNO₃ + 10K = 6K₂O + N₂

n(K)=m(K)/M(K)
n(K)=155g/39.1g/mol=3.964 mol

n(KNO₃)=m(KNO₃)/M(KNO₃)
n(KNO₃)=122g/101.1g/mol=1.207 mol

K:KNO₃ = 10:2 =5:1
3.964:1.207=3.3:1

K is the limiting reagent

Answer 2

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