I think it’s 2.46 moles of H2O, unless the ‘burns in oxygen’ is relevant to the equation. i hope this helps.
64.0g C2H2 X 1 mol C2H2 over 26.04g C2H2 X 2 mol H2O over 2 mol C2H2
The first step of the equation is a basic diversion factor showing that for ever C2H2, there’s one mole. I got the number 26.04 grams of C2H2 by calculating their combined mass. To do that, I added them from the periodic table for and multiplied it by the amount of each that I had. Which was displayed as 12.01(2) + 1.01(2)
The second step to the equation is the mole conversion factor. In this step, I took the number of moles from the equation. That just shows that for every 2 moles of C2H2, there will be 2 moles of H2O.
After setting it up, you can ignore any ones in the problem as they won’t make any difference in the answer. And since, in this equation, the moles in the mole conversion factor are equal, you may ignore those too as they will divide down to one and not have any significance.
Now, to find the answer I divided 64.0g C2H2 by 26.04g C2H2 since everything else cancelled out. It equals 2.457757296 moles of H20. But, to get get credit for the sig-figs, you’ll have to change the numbers before submitting.
The beginning number has three placements (64.0) so the answer must also have three. If we look at the long decimal we got in the calculator, we can go over three and round the last number either up or down depending on the fourth number.
