Question

write a balanced chemical equation based on the following description: sulfur dioxide gas reacts with solid potassium telluride to produce sulfur ditelluride gas and solid potassium oxide 2so₂(g) + 4k₂te(s) → 2s₂te₂(g) + 4k₂o(s)

Explanation:

Step1: Write the un - balanced equation

Sulfur dioxide (\(SO_2\)), potassium telluride (\(K_2Te\)), sulfur ditelluride (\(S_2Te_2\)) and potassium oxide (\(K_2O\)). The un - balanced equation is \(SO_2(g)+K_2Te(s)\to S_2Te_2(g)+K_2O(s)\)

Step2: Balance the sulfur atoms

On the left - hand side, there is 1 \(S\) atom in \(SO_2\), and on the right - hand side, there are 2 \(S\) atoms in \(S_2Te_2\). So we put a coefficient of 2 in front of \(SO_2\): \(2SO_2(g)+K_2Te(s)\to S_2Te_2(g)+K_2O(s)\)

Step3: Balance the tellurium atoms

On the left - hand side, there is 1 \(Te\) atom in \(K_2Te\), and on the right - hand side, there are 2 \(Te\) atoms in \(S_2Te_2\). So we put a coefficient of 2 in front of \(K_2Te\): \(2SO_2(g)+2K_2Te(s)\to S_2Te_2(g)+K_2O(s)\)

Step4: Balance the potassium atoms

On the left - hand side, there are 4 \(K\) atoms (\(2\times2\) from \(2K_2Te\)), so we put a coefficient of 2 in front of \(K_2O\): \(2SO_2(g)+2K_2Te(s)\to S_2Te_2(g)+2K_2O(s)\)

Step5: Check the balance of all atoms

• Sulfur: Left - hand side \(2\) (from \(2SO_2\)), right - hand side \(2\) (from \(S_2Te_2\))
• Oxygen: Left - hand side \(4\) (from \(2SO_2\)), right - hand side \(2\) (from \(2K_2O\)). So we need to adjust.

Let's start over.
Let the balanced equation be \(aSO_2 + bK_2Te=cS_2Te_2 + dK_2O\)
For \(S\): \(a = 2c\)
For \(Te\): \(b = 2c\)
For \(K\): \(2b=2d\) (or \(b = d\))
For \(O\): \(2a=d\)
Let \(c = 1\), then \(a = 2\), \(b = 2\), \(d = 2\)

Answer:

\(2SO_2(g)+2K_2Te(s)\to S_2Te_2(g)+2K_2O(s)\)