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What is the ratio volume of volume occupied by 1 mole of o2 and 1 mole of o3?
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How many moles of H2 are required to react with 5.0 moles O 2?
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.
How many moles of P2 O5 are produced from 8.00 mole of O2?
Balanced equation first. 4P + 5O2 -> 2P2O5 8.00 mole O2 (2 mole P2O5/5 mole O2) = 3.20 moles P2O5 produced
How do you figure out 3.2 moles of o2?
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 --------------------------------------------------
What is the difference between the mass of a mole of oxygen and a mole of O2 molecules?
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.
How many moles O2 must react with excess H2 to form 7.89 mol H2O?
Oxygen limits the reaction, so......Balanced equation. 2H2 + O2 -> 2H2O 7.89 mole H2O (1 mole O2/2 mole H2O) = 3.95 mole oxygen gas needed ------------------------------------------
Will 27 grams of aluminium react completely with 24 grams of oxygen?
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
At STP 32 grams of O2 would occupy the same volume as?
PV=nRT 32 gram O2 = 1 mole O2 (1atm)(V) = (1 mole)(.0821)(273) V = 22.4 L
How many moles of H2 are required to react with 5.0 moles O 2?
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.
How much space does one mole of gas take?
At STP it is about 22.44 liters, but I can never remember this, so the ideal gas equation to back up this answer. pressure*volume = moles times a constant* temperature in Kelvin PV = nRT (1 atm)(volume) = (1 mole O2)(0.08206 L*atm/mol*K)(298.15 K) Volume = 24.47 Liters ( space occupied ) -----------------------------------------------------------
What amount of oxygen require to react one gram calcium?
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
If 0.440 mole of HgO reacts how many moles of O2 will be produced?
0.220 mole O2
Best Hydrogen gas to Oxygen gas ratio for combustion to occur?
The reaction is 2H2 + O2-> 2H2O So, both the molar and volume ratios would be 2 moles(liters) of hydrogen to every mole(liter) of oxygen. By weight, the ratio would be about 8 grams of oxygen to every gram of hydrogen
How many moles of P2 O5 are produced from 8.00 mole of O2?
Balanced equation first. 4P + 5O2 -> 2P2O5 8.00 mole O2 (2 mole P2O5/5 mole O2) = 3.20 moles P2O5 produced
The number of molecules in 48.0 grams of oxygen gas (O2) is?
48.0 g O2 x 1 mole O2/32 g x 6.02x10^23 molecules O2/mole O2 = 9.03x10^23 molecules of O2
What is the volume of oxygen required for complete combustion of 0.5 liter of H2S at stp?
For gases it is valid that the Volume ratio of reactants and products is the same as the mole ratio (in the balanced equation) when pressure and Temperature are kept coonstant. This is according to the general gas law (Boyle-Gay-Lussac): p.V = m.R.T2H2S + 3O2 --> 2H2O + 2SO2so 0.5 L H2S needs 0.5 * [3/2] = 0.75 L O2
What happens if 0.1 mole methane reacts with 0.1 mole oxygen gas?
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.)
When 1.20 grams of carbon react with 2.24 Liters of oxygen gas at Standard Temperature Pressure how many moles of carbon monoxide are produced?
ok well standard temperature and pressure is stp 22.4 litres per mole of gas. first write a balanced chemical equation 2C + 02 ------> 2CO then write mole ratios 2C + 02 ------> 2CO 2 : 1 : 2 using gay lussacs law and avogadros hypothesis we know that the same volumes of gases in same conditions are mole ratios (only works with gases however) so therefore the volume of CO gas produced will be 2 x that of the volume of oxygen gas (because ratio of O2 gas to CO gas is 1:2) =2x2.24 =4.48 litres of CO gas produced then the moles can be worked out by - number of moles = volume given/volume of 1 mole at STP = 4.48/22.4 =0.2 moles roughly i think. Answer: 1.20 g C is 1/10 mole of C 2.24 liters of O2 is 1/10 mole of O2So when equal volumes molar anounts of C and O2 react: C (1 mole)+ O2(1 mole)-> CO2 (1 mole) (No CO is produced) With the given initial amounts 1/10 mole (2.24 liters) of CO2 would be producedIf oxygen were limited all C could be converted to CO. This would rturn the inital amount of C (1,2 g) and 1.12 liters of O2 into 1/10 mole of CO (2.24 liters)
