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1mol of a gas occupies 24 dm3 at STP, so
2.2mol X 24 mol/dm3
=52.8dm3 or 5280cm3
Wiki User
∙ 14y ago
Wiki User
∙ 11y ago1 mole of He occupies 22.414 liters at STP.
So, 2.1 mole will occupy 502.4 liters
Wiki User
∙ 11y ago1 mole occupies 22.4 liters.
So 0.20 moles will occupy 4.48 liters
Wiki User
∙ 8y ago2.4mol x 22.4 = 54L
Wiki User
∙ 8y agoThe volume is 56,5 L.
Wiki User
∙ 12y ago49.7 l o2
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What volume of O2 at stp is necessary to react with o200 moles of SO2 to form SO3?
2.24 L O2 (= 0.100 mol O2) is needed to react with 0.200 moles of SO2 to form SO3
Does oxygen and nitrogen weigh the same in volume?
No, one mole of each, having the same VOLUME (about 22.4 L at STP), differ though in their masses: 32 g/mol for O2 and 28 g/mol for N2 So their densities (mass per volume) also differ in the same way: 1.43 g/L and 1.25 g/L respectively, at STP.
Does nitrogen and oxygen weigh the same?
No, one mole of each, having the same VOLUME (about 22.4 L at STP), differ though in their masses: 32 g/mol for O2 and 28 g/mol for N2 So their densities (mass per volume) also differ in the same way: 1.43 g/L and 1.25 g/L respectively, at STP.
What is the volume of oxygen that will be used by 12g of Mg to be completely converted into MgO?
The reaction : Mg + O2-----------> MgO The balanced reaction : 2 Mg + O2-----------> 2 MgO that is 2 moles of Mg reacts with one mole of oxygen gas. moles of Mg reacted = 12 g / 24 g mol-1=0.5 mol. Thus the no. of mol of O2 reacted = 0.5 mol /2 =0.25 mol If the gas is at standard temperature and pressure, then Volume of 1 mol of O2 gas = 22.4 dm3 Volume of 0.25 mol of O2 gas = 22.4 x 0.25 dm3= 5.6 dm3 the volume of oxygen that will be used to react completely with 12 g Mg =5.6 dm3 (provided that the gas is at the standard conditions.)
How many liters of pure oxygen at STP is consumed by a human being in 24 hours if the human body requires daily energy that comes from metabolizing 816 grams of sucrose c12 h22 o11?
816 g C12H22O11 x (1 mol C12H22O11 / 342.0 g C12H22O11) x (12 mol O2 / 1 mol C12H22011) x (22.4 L O2 / 1 mol O2) = 641 L O2
How many liters of oxygen gas are present in 11.3 grams of oxygen gas?
Because oxygen gas (O2) has a molar mass of 32g/mol, 11.3 g * 1/32 mol/g gives about .35 moles. An ideal gas has a volume of 22.4 L/mol at STP, so 11.3 g O2 would have a volume of 7.91 L at STP.
What is the density O2 mol at STP?
1.43 g/cm3
What volume of O2 at stp is necessary to react with o200 moles of SO2 to form SO3?
2.24 L O2 (= 0.100 mol O2) is needed to react with 0.200 moles of SO2 to form SO3
Does oxygen and nitrogen weigh the same in volume?
No, one mole of each, having the same VOLUME (about 22.4 L at STP), differ though in their masses: 32 g/mol for O2 and 28 g/mol for N2 So their densities (mass per volume) also differ in the same way: 1.43 g/L and 1.25 g/L respectively, at STP.
Does nitrogen and oxygen weigh the same?
No, one mole of each, having the same VOLUME (about 22.4 L at STP), differ though in their masses: 32 g/mol for O2 and 28 g/mol for N2 So their densities (mass per volume) also differ in the same way: 1.43 g/L and 1.25 g/L respectively, at STP.
What is the volume of oxygen that will be used by 12g of Mg to be completely converted into MgO?
The reaction : Mg + O2-----------> MgO The balanced reaction : 2 Mg + O2-----------> 2 MgO that is 2 moles of Mg reacts with one mole of oxygen gas. moles of Mg reacted = 12 g / 24 g mol-1=0.5 mol. Thus the no. of mol of O2 reacted = 0.5 mol /2 =0.25 mol If the gas is at standard temperature and pressure, then Volume of 1 mol of O2 gas = 22.4 dm3 Volume of 0.25 mol of O2 gas = 22.4 x 0.25 dm3= 5.6 dm3 the volume of oxygen that will be used to react completely with 12 g Mg =5.6 dm3 (provided that the gas is at the standard conditions.)
How many liters of pure oxygen at STP is consumed by a human being in 24 hours if the human body requires daily energy that comes from metabolizing 816 grams of sucrose c12 h22 o11?
816 g C12H22O11 x (1 mol C12H22O11 / 342.0 g C12H22O11) x (12 mol O2 / 1 mol C12H22011) x (22.4 L O2 / 1 mol O2) = 641 L O2
How many milliliters of O2 will form a STP from 55.2 grams of KCLO3?
I assume you mean the decomposition reaction used to produce O2 in lab. 2KCLO3 -> 2KCl + 3O2 find moles O2 55.2 grams KClO3 (1 mole KCLO3/122.55 grams)(3 mole O2/2 mole KClO3) = 0.67564 moles O2 Now, I use the ideal gas law PV = nRT (1 atm)(V) = (0.67564 mol)(0.08206 L*atm/mol*K)(298.15 K) Volume O2 = 16.53 Liters which is 16530 milliliters ( less significant figures )
How many atoms are present in 16g of oxygen gas?
See the Related Question linked the the left of this answer: = How do you solve an Ideal Gas Law problem? = Note that STP is standard temperature and pressure. Standard pressure is 1 atm, and standard temperature is 0 °C, which is 273.15 Kelvin.
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
Calculate the volume of oxygen gas required at 1.75 ATM and 21C to burn completely C6H6 to water vapor and 2.56 L of carbon dioxide gas at 84.7 kPa and 34C?
Write a chemical equation in symbols for this reaction.Write a balanced chemical equation.What does the problem ask you to find? Volume of oxygen gas at 1.75 atm and 21°C. Write this on the right side.What does the problem give you to start with? 2.56 L of carbon dioxide gas at 84.7 kPa and 34°C. Write this on the left side.Change the given volume from the non-standard conditions to STP using the combined gas laws.Convert from liters to moles using 22.4 L/mol. (You can use 22.4 L/mol because you corrected the given volume to STP.)Convert from mol CO2 to mol O2 . Use the coefficients from the balanced equation.Convert from mol O2 to liters of O2 at STP.Change the volume of oxygen from STP to the requested conditions of 84.7 kPa and 34°C using the combined gas laws. Make certain that your pressure units are the same. If not, then you must convert one to the other. Cancel all units and calculate using the procedures that you have been taught. This problem has a lot of numbers to enter. You should always calculate a problem like this twice to make certain that you get the same answer each time.You should make note that if done properly the standard temperatures and pressures on each end will cancel. Since this is a volume-to-volume problem, then 22.4 will cancel.Answer: 472 g Mg3 (PO4)2
How many atoms are in 12.54 g of O2?
12.54 (g O2) / 2*15.99 (g/mol O2) = 0.3921 mol O2 -->0.3921 (mol O2) * 6.022*1023 (molecules O2)/(molO2) == 2.361*1023 molecules O2= 4.723*1023 atoms O
