Hydrogen sulfide gas burns in the presence of oxygen to produce sulfur dioxide gas and water vapor. Write a balanced chemical equation, including the physical state

2H2S + 3O2 --> 2SO2 + 2H2O If this doesn't help, then I have no idea

The problem asks to show the states, also. The equation is correct.

2H2S(g) + 3O2(g) --> 2SO2(g) + 2H2O(g)
Note the problem TELLS you the water is vapor so you know to write it as a gas and not a liquid.

To write a balanced chemical equation for the burning of hydrogen sulfide gas in the presence of oxygen, we need to follow a few steps:

1. Identify the reactants and products: The reactants are hydrogen sulfide gas (H2S) and oxygen gas (O2), while the products are sulfur dioxide gas (SO2) and water vapor (H2O).

2. Write the unbalanced chemical equation: H2S + O2 → SO2 + H2O

3. Balance the equation: Start by balancing the element that appears in the least number of compounds on either side of the equation, which is sulfur (S) in this case. There is one sulfur atom on each side, so we don't need to balance it further.

Next, balance the hydrogen (H) atoms by adding a coefficient of 2 in front of the water vapor on the product side:

H2S + O2 → SO2 + 2H2O

Now, let's balance the oxygen (O) atoms. We have 2 oxygen atoms on the left side from the oxygen gas molecule and 2 oxygen atoms on the right side from the sulfur dioxide molecule. Adding these together, we have a total of 4 oxygen atoms on the right side. To balance this, we need to add a coefficient of 2 in front of the oxygen gas on the reactant side:

2H2S + 2O2 → 2SO2 + 2H2O

Finally, check that all the atoms are balanced. On the left side, we have 4 hydrogen atoms (2 from each H2S) and 4 oxygen atoms (2 from each O2). Similarly, on the right side, we have 4 hydrogen atoms (2 from each H2O) and 4 oxygen atoms (2 from each SO2), which confirms that the equation is balanced.

The balanced chemical equation for the burning of hydrogen sulfide gas in the presence of oxygen is:

2H2S(g) + 2O2(g) → 2SO2(g) + 2H2O(g)