Question

when balancing this equation like the one above, you may wonder where to start. there are three reactants and three unique products. here are two methods you can use to help determine where to start. using the first method, start with kno₃, lets first look at k or potassium atoms on both sides of the equation. on the products there are two k atoms, so to match this, well place a 2 before potassium nitrate. balance the equation below: given this partially balanced equation, its up to you to balance the rest. before submitting your answer, make sure to fill in all number inputs (put a 1 if you dont intend to add a coefficient). 2 kno₃ + al + s → k₂s + n₂ + al₂o₃

Explanation:

Step1: Balance K atoms

Given \(2KNO_3\), on the product side, we have \(K_2S\). Since there are 2 K atoms on the reactant side (from \(2KNO_3\)), the coefficient of \(K_2S\) is 1.

Step2: Balance N atoms

From \(2KNO_3\), there are 2 N atoms. On the product side, \(N_2\) has 2 N atoms. So the coefficient of \(N_2\) is 1.

Step3: Balance O atoms

From \(2KNO_3\), there are \(2\times3 = 6\) O atoms. On the product side, \(Al_2O_3\) has O atoms. Let the coefficient of \(Al_2O_3\) be \(x\). Then \(3x=6\), so \(x = 2\).

Step4: Balance Al atoms

Since the coefficient of \(Al_2O_3\) is 2, there are \(2\times2=4\) Al atoms on the product side. So the coefficient of Al on the reactant side is 4.

Step5: Balance S atoms

Since the coefficient of \(K_2S\) is 1, there is 1 S atom on the product side. So the coefficient of S on the reactant side is 1.

Answer:

\(2KNO_3+4Al + 1S
ightarrow1K_2S+1N_2 + 2Al_2O_3\)