A sample of 4.00 g of methane os mixed with15.0 grams of chlorine determine the limiting reactant according to the chem equa : CH4+4Cl2->CCl4+4HCl
what is the max mass of CCl4 that can be formed
2006-10-24 17:25:07 · 3 answers · asked by coolpuffin 2 in Science & Mathematics ➔ Chemistry
One good thing about chemical equations is that no matter what unit of measurement is used (mole, grams, atomic mass,...) the ratios will always stay the same. for example, if you have:
2H2 + O2 -------> 2H2O
you can say for every 2 molecules of hydrogen you need 1 molecule of oxygen to form 2 molecules of water.
Or you can say for 2 moles of hydrogen you need 1 mole of oxygen to form 1 mole of water; and so on.
So your ratio is 2:1:2
From your equation: CH4 + 4Cl2 -----> CCl4 + 4HCl
your ratio is 1:4:1:4
So 1 gram of CH4 reacts with 4 grams of Cl2 to produce 1 gram of CCl4 and 4 grams of HCl.
Therefore 4 grams of CH4 reacts with 16 grams of Cl2. So if you only have 15 grams of Cl2, your Cl2 will be your limiting factor and as soon as you run out of Cl2, your equation will stop and you will have some CH4 left.
2006-10-24 17:45:50 · answer #1 · answered by smarties 6 · 0⤊ 0⤋
Convert both in to moles. Look at ur mole ratios 2 see if u have enough of the other compound to react. If for example when run through with the methane u see the mole ratio says u need more chlorine than u actually have than chlorine is ur limiting reagent. From there run through the equation using that compound to find how much ccl u get. easy right ;)
2006-10-24 17:32:17 · answer #2 · answered by Blahblah_bbbllaah 2 · 1⤊ 1⤋
4.00g CH4 (1 mol CH4/ 16.04gCH4)(1mol CCl4/1 molCH4)(154gCCL4/1 mol CCl4) = 38.4 g CCl4
15.0g Cl2 (1 mol Cl2/ 71 g Cl2)(1 mol CCl4/ 4 mol Cl2)(154gCCl4/1 mol CCl4) = 8.13 g CCl4 = Limiting Reactant & Maximum Mass of CCl4
2006-10-24 17:47:06 · answer #3 · answered by Ad Just 4 · 0⤊ 0⤋