You use the Mole-to-Mole ratio. If the equation is 2CH4 + 2H2O = 6H2 + 2CO, then you would start with your given, 8.0 mol CO and multiply that with your mol-to-mol ratio which is (2mol CO/ 2 mol CH4). Your answer will be 8.0 mol.
the experimental mole ratio has a bigger penis
Balanced equation first. 4P + 5O2 -> 2P2O5 8.00 mole O2 (2 mole P2O5/5 mole O2) = 3.20 moles P2O5 produced
The mass of 3.2 moles O2? The atoms in 3.2 moles O2 Could be other things, I guess, but I will do these two. 3.2 moles O2 (32 grams/1 mole O2) = 102.4 grams of gaseous oxygen -------------------------------------------- 3.2 mole O2 (6.022 X 1023/1 mole O2) = 1.9 X 1024 atoms of gaseous oxygen --------------------------------------------------
A mole of oxygen atoms has a mass of approximately 16 grams. A mole of O2 has a mass of approximately 32 grams. A mole is 6.02 x 1023 particles and as such a mole of oxygen atoms has only half the mass of a mole of oxygen molecules.
Yes.Explanationary:27 g Al = 1.0 mole Al24 g = 1.5 mole O2 so this ratio (in mole) is 1:1.52Al + 3O2 --> Al2O3 so the balanced mole ratio is 2:3 or 1:1.5
the experimental mole ratio has a bigger penis
You use the Mole-to-Mole ratio. If the equation is 2CH4 + 2H2O = 6H2 + 2CO, then you would start with your given, 8.0 mol CO and multiply that with your mol-to-mol ratio which is (2mol CO/ 2 mol CH4). Your answer will be 8.0 mol.
This is a relatively simple question that can be answered using simple stochiometry. If we assume a few things: a) that pure elements are reacting, b) that oxygen is in its stable diatomic form, and c) that they are reacting on a 1 to 1 mole ratio; then the math is as follows: 1 g.O2 x (1 mole O2 / 32 grams per mole) x (1 mole Ca / 1 mole O2) x (40.078 grams per mole / 1 mole Ca) = XXX.XX grams of Ca Thus the amount of calcium that will react on a 1 to 1 mole ratio with one gram of diatomic oxygen (O2) is 1.252 grams of calcium
0.220 mole O2
Balanced equation first. 4P + 5O2 -> 2P2O5 8.00 mole O2 (2 mole P2O5/5 mole O2) = 3.20 moles P2O5 produced
48.0 g O2 x 1 mole O2/32 g x 6.02x10^23 molecules O2/mole O2 = 9.03x10^23 molecules of O2
C4H9OH + O2 → CO2 + H2O (Unbalanced)C4H9OH + 6O2 → 4CO2 + 5H2O (Balanced)The 'balanced' equation above is not correct, however the one below is!2C4H9OH + 12O2 → 8CO2 + 10H2O
Lets start with the reaction of methane fully reacting with oxygen: CH4 + 2 O2 --> CO2 + 2 H2O The ratio CH4:O2 is 1:2 So 0.1 mole of methane can potentially react with 0.2 mole of oxygen. Seeing as we only have 0.1 mole oxygen we know that we have excess of methane. Because of the lack of oxygen the methane can form carbonmonoxide (CO) instead of CO2. (if someone knows the composition of the final product, please add here. As far as i know, you cant predict the CO:CO2 ratio.)
The mass of 3.2 moles O2? The atoms in 3.2 moles O2 Could be other things, I guess, but I will do these two. 3.2 moles O2 (32 grams/1 mole O2) = 102.4 grams of gaseous oxygen -------------------------------------------- 3.2 mole O2 (6.022 X 1023/1 mole O2) = 1.9 X 1024 atoms of gaseous oxygen --------------------------------------------------
4NH3 + 5O2 -> 4NO + 6H2O I suspect NH3 limits. Let's see. 5.15 O2 ( 4 mole NH3/5 mole O2) = 4.12 mole NH3 you do not have that much ammonia, so it limits and drives the reaction. 3.80 mole NH3 (4 mole NO/4 mole NH3) = 3.80 moles of NO made
A mole of oxygen atoms has a mass of approximately 16 grams. A mole of O2 has a mass of approximately 32 grams. A mole is 6.02 x 1023 particles and as such a mole of oxygen atoms has only half the mass of a mole of oxygen molecules.