Atoms of three different elements are represented by \(\mathrm{O}, \square,\) and \(\Delta .\) Which compound is left over when three molecules of \(\mathrm{O}\Delta\) and three molecules of \(\square \square \Delta\) react to form \(O \square \Delta\) and \(\mathrm{O} \Delta \Delta ?\)

There are three molecules of OΔ and three molecules of □□Δ. Count the number of each element in each molecule and sum them together. - OΔ contains 1 O and 1 Δ, so three molecules of OΔ contain 3 O's and 3 Δ's. - □□Δ contains 2 □ and 1 Δ, so three molecules of □□Δ contain 6 □'s and 3 Δ's. Before the reaction, there are 3 O's, 6 □'s, and 6 Δ's in total.

The reaction forms O□Δ and OΔΔ. We need to find out how many of each element is present in these two compounds. - O□Δ contains 1 O, 1 □, and 1 Δ - OΔΔ contains 1 O and 2 Δ's Now let's assume there is "x" molecule of O□Δ and "y" molecule of OΔΔ after the reaction. The total number of atoms after the reaction will be: - O's: 1x + 1y = x + y - □'s: 1x = x - Δ's: 1x + 2y = x + 2y

Now we need to compare the number of atoms before and after the reaction. We know that the number of atoms is conserved, meaning the number of atoms before the reaction should be equal to the number of atoms after the reaction. 1. For O's: 3 (before) = x + y (after) 2. For □'s: 6 (before) = x (after) 3. For Δ's: 6 (before) = x + 2y (after) Using the second equation (for □'s), we find that x = 6. This means that there are 6 molecules of O□Δ formed in the reaction. Now we have to find the value of y by using the equations for O's and Δ's. If we substitute x = 6 into the first equation (for O's), we get: 3 = 6 + y y = -3 Since the value of y is -3, this means that there are no Δ's left to form any OΔΔ molecule. So, the compound left over is OΔ (3 OΔ molecules to be precise), as OΔΔ couldn't be formed due to the lack of Δ atoms.