figure how many moles of Sb is one gram.
Then, you need 1/2 that number of moles of Sb2O4. That is the number of moles of Sb2O4 in the first reaction. Then you need 5 times that number of moles of O2.
putting it together, you figure the moles of Sb. Then you need 5/2 of that of oxygen for the first equation.
Now on the second q, it is totally different. You will have to figure the limiting reactant in the equation in the reaction. Figure the moles of Sb2S3, and O2 from the masses given. Then, do you have 5 times as much O2? If you dont, then oxygen is limiting, and you figure the reaction based on just the moles of oxgen. If you have more than 5x , then Sb2S2 is limiting, and you figure the reaction based on that amount.
Elemental antiomony made by these reactions down below:
Sb2S3 + 5 O2 - > Sb2O4 + 3 SO2
Sb2O4 + 4 C - > 2 Sb + 4 CO
how many grams O2 needed to prepare 1.00g of Sb?
how many grams Sb formed when 84.9g of Sb2S3 and 39.0g of O2 r used in first reaction and product formed is used in second reaction?
i not understand these.
3 answers
i actually doing this first time and i real not get how to do this how i find out how many mole Sb in one gram?
moles = grams/molar mass