Enough concentrated sulfuric acid solution was added to provide 3 grams of sulfuric acid which dissolved i4 grams of pure fluorapatite. How much excess sulfuric acid was used? (report answer in grams of H2SO4
A poorly worded question. What you want to know is how much H2SO4 was used, not how much xs was used. Technically, none of the excess was used. Did your problem provide an equation. Check and see that it looks like this.
Ca5(PO4)3F + 5H2SO4 ==> 5CaSO4 + 3H3PO4 + HF
If my equation is right 3 g H2SO4 will not dissolve 14 g of fluoroapatite.
To find out how much excess sulfuric acid was used, we need to determine the exact amount of sulfuric acid that reacted with the fluorapatite.
Given that 3 grams of sulfuric acid reacted with 14 grams of fluorapatite, we can set up a ratio:
3 grams H2SO4 / 14 grams fluorapatite = x grams H2SO4 / 4 grams fluorapatite
Here, x represents the amount of sulfuric acid we want to find.
To solve for x, we can cross-multiply and then divide:
3 grams H2SO4 * 4 grams fluorapatite = 14 grams fluorapatite * x grams H2SO4
12 grams H2SO4 = 14 grams fluorapatite * x grams H2SO4
x = 12 grams H2SO4 / 14 grams fluorapatite
x ≈ 0.857 grams H2SO4
Therefore, approximately 0.857 grams of excess sulfuric acid was used.