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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.)
What is the volume of 0.24 mol oxygen gas at 345 k and a pressure of 0.75 ATM?
The volume is 9,06 L.
When 46.0 g KClO3 is heated 12.1 g O2 is formed how many grams of KCl is also formed?
The equation that describes this process is as follows: 2KClO3 ---> 2KCl + 3O2 For every 2 moles of reactants consumed 3 moles of oxygen gas are produced. 3 mol O2 / 2 mol KClO3 = x mol O2 / 12.3 mol KClO3 x = 12.3 mol x 3 mol / 2 mol = 18.45 mol Therefore, 18.5 mol (3 significant figures) of oxygen are produced by the decomposition of 12.3 mol of potassium chlorate
How do you calculate the mass of aluminum oxide produced when 3.75 moles of aluminum burn in oxygen?
Aluminum Oxide is Al2O3 and Al = 27 and oxygen =16 so the molar mass is 102 g/mol Al2O33.75 mol Al ~ 3.75/2 mol Al2O3 ~ (3.75/2)mol * 102 g/mol = 191.25 = 191 gram Al2O3
How many moles of oxygen atoms are in 4.25 moles of calcium carbonate the chief constituent of seashells?
Calcium Carbonate (CaCO3)I mol CaCO3 contains 3 mol Oxygen atomsso 4.25 mol CaCO3 will have 12.75 mol Oxygen Atoms.
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.)
What is the volume of 0.24 mol oxygen gas at 345 k and a pressure of 0.75 ATM?
The volume is 9,06 L.
How do you figure out the HNO3 Concentration?
The equation you will need is: Mol of substance 1 * volume of substance 1 = Mol of substance 2 * volume of substance 2
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 mass and no of molecules of O2 having volume 22.4 dm3?
22,4 L is the molar volume. 1 mol of oxygen has 32 g and and 6,022140857.10e23 molecules.
When 46.0 g KClO3 is heated 12.1 g O2 is formed how many grams of KCl is also formed?
The equation that describes this process is as follows: 2KClO3 ---> 2KCl + 3O2 For every 2 moles of reactants consumed 3 moles of oxygen gas are produced. 3 mol O2 / 2 mol KClO3 = x mol O2 / 12.3 mol KClO3 x = 12.3 mol x 3 mol / 2 mol = 18.45 mol Therefore, 18.5 mol (3 significant figures) of oxygen are produced by the decomposition of 12.3 mol of potassium chlorate
How do you calculate the mass of aluminum oxide produced when 3.75 moles of aluminum burn in oxygen?
Aluminum Oxide is Al2O3 and Al = 27 and oxygen =16 so the molar mass is 102 g/mol Al2O33.75 mol Al ~ 3.75/2 mol Al2O3 ~ (3.75/2)mol * 102 g/mol = 191.25 = 191 gram Al2O3
How do you calculate volume in percentage?
it's the same as the molar percentage as 1 mol of gas will occupy 24.79dm3/mol at 25 °C no mater what the element/compound is so if you know the actual amounts calculate the molar fraction and that will give you the percentage
Which gas is heavier oxygen or nitrogen?
Since both form diatomic elements, we simply have to compare molecular masses. O2 has a mass 32.0 g/mol, while N2 is 28.0 g/mol. This means that, since gases (according to the ideal gas law) all contain 22.4 mol/L, then the same volume of oxygen would be heavier than the same volume of nitrogen.
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.
What is the volume of 10.9 mol of helium at STP?
The volume of 10.9 mol of helium at STP is 50 litres.
How many molecules of water can be produced by the reaction of 2.16 mol of oxygen with excess Ethylene.?
C2H4 + 3 O2 --> 2 CO2 + 2 H2OSo 2.16 mol O2 will produce 1.44 mol H2O(and 1.44 mol CO2)because 3:2 = 2.16 : 1.44
