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What volume (in milliliters) of oxygen gas is required to react with 4.03 g of Mg at STP?
2Mg + O2 ==> 2MgO Balanced Equation4.03 g Mg x 1 mole Mg/24.3 g = 0.166 moles Mg present in 4.03 g Mg.moles O2 required = 0.166 moles Mg x 1 mole O2/2 moles Mg = 0.083 moles O2 needed.At STP 1 mole occupies 22.4 L, thusVolume of O2 required = 0.083 moles x 22.4 L/mole = 1.56 L x1000 ml/L = 1859 mlsSince 4.03 g has only 3 significant figures, the correct answer should be 1860 milliliters.
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
What volume of oxygen at STP can be profuced by 6.125g of potassium chlorate according to the reaction?
The balanced reaction for the decomposition of potassium chlorate (KClO3) is: 2 KClO3 -> 2 KCl + 3 O2 From the reaction, 2 moles of KClO3 produce 3 moles of O2. Calculate the moles of KClO3 in 6.125g using its molar mass. Convert moles of KClO3 to moles of O2. Use the ideal gas law to find the volume of O2 at STP (22.4 L/mol).
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.
What volume of O2 at stp is necessary to react with o200 moles of SO2 to form SO3?
The balanced equation for the reaction is: 2SO2 + O2 -> 2SO3. Therefore, 1 mole of O2 is needed to react with 2 moles of SO2 to form 2 moles of SO3. So for 200 moles of SO2, you would need 100 moles of O2. At STP, 1 mole of any gas occupies 22.4 L, so the volume of O2 needed would be 2240 L (100 moles x 22.4 L).
What is the density O2 mol at STP?
1.43 g/cm3
What volume (in milliliters) of oxygen gas is required to react with 4.03 g of Mg at STP?
2Mg + O2 ==> 2MgO Balanced Equation4.03 g Mg x 1 mole Mg/24.3 g = 0.166 moles Mg present in 4.03 g Mg.moles O2 required = 0.166 moles Mg x 1 mole O2/2 moles Mg = 0.083 moles O2 needed.At STP 1 mole occupies 22.4 L, thusVolume of O2 required = 0.083 moles x 22.4 L/mole = 1.56 L x1000 ml/L = 1859 mlsSince 4.03 g has only 3 significant figures, the correct answer should be 1860 milliliters.
What volume of oxygen is needed to react with a solid sulfur to form 3.5L of SO2?
1 mole of gas occupies 22.4 liters at STP. Therefore 3.5/22.4 = 0.15625 moles of SO2. There are thus 0.15625 moles of O2 needed to react with solid sulfur because S + O2 ---->SO2. 0.15625 moles of oxygen occupies 0.15625 x 22.4 liters = 3.5 liters O2 required.
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
What volume of oxygen at STP can be profuced by 6.125g of potassium chlorate according to the reaction?
The balanced reaction for the decomposition of potassium chlorate (KClO3) is: 2 KClO3 -> 2 KCl + 3 O2 From the reaction, 2 moles of KClO3 produce 3 moles of O2. Calculate the moles of KClO3 in 6.125g using its molar mass. Convert moles of KClO3 to moles of O2. Use the ideal gas law to find the volume of O2 at STP (22.4 L/mol).
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.
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.
What volume of O2 at stp is necessary to react with o200 moles of SO2 to form SO3?
The balanced equation for the reaction is: 2SO2 + O2 -> 2SO3. Therefore, 1 mole of O2 is needed to react with 2 moles of SO2 to form 2 moles of SO3. So for 200 moles of SO2, you would need 100 moles of O2. At STP, 1 mole of any gas occupies 22.4 L, so the volume of O2 needed would be 2240 L (100 moles x 22.4 L).
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 grams would 2350 L of O2 gas at STP weigh?
To find the weight of 2350 L of O2 gas at STP, you would first need to calculate the moles of gas using the ideal gas law. Then, use the molar mass of O2 to convert moles to grams. The molar mass of O2 is 32 g/mol, so you would multiply the moles by 32 g/mol to find the weight in grams.
How many moles of oxygen in 30L of oxygen gas?
At standard temperature and pressure (STP), one mole of a gas is 22.4L. So, in order to determine how many moles of O2 are in 30L, you do the following: multiply 30L O2 x 1mol O2/22.4L O2, which equals 1.34mol O2.
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
