3 posts • Page 1 of 1
A reaction vessel contains 5.77g of white phosphorus and 5.77g of oxygen. The first reaction to take place is the formation of phosphorus (III) oxide, P406:P4(s) + 302 (g) ---> P4O10 (s). (a) What is the limiting reactant for the formation of P4O10? (b) What mass of P4O10 is produced? (c) How many grams of the excess reactant remain in the reaction vessel?
Hey there, wow this was quite the question. So basically, you take the masses you are given of P4 and O2 and convert them to moles by dividing by their respective molar masses. You also compare the P4 moles to the O2 moles by multiplying the P4 moles by 3 since the ratio of P4 to O2 atoms is 1:3. Then, you will see that this amount of O2 moles is less than the amount of O2 moles you were given originally, meaning you have excess O2, which makes P4 the limiting reactant. You then take the P4 moles and multiply it by the molar mass of the product to see how much product you have. Going back to excess O2, you subtract the O2 moles you used by the O2 moles you were given, to get the extra O2 moles which will move on to the next reaction. Hope this helps :)
You can calculate that you have 0.4657mol P4 while have 0.18 mole O2, which means O2 is in excess. So O2 is the limiting reagent. So the P4 will be consumed up and you would get 0.04657 mole P4O6. In the second reaction, you only have 0.036 mole O2 left, which is the limiting reagent. And you could only get about 0.018 mole P4O10, and calculate it into mass. And then you can find how many P4O6 left and calculate it to mass. Hope this would help.
Who is online
Users browsing this forum: No registered users and 1 guest