what is ag2o
Balanced Formula:2Mg + O2 --> 2MgOMole ratio:2 : 1 : 2Givens:.486 g oxygen.738 g magnesium24.3 g = atomic mass of magnesium16.0 g = atomic mass of oxygen40.3 g = molecular mass of magnesium oxideFind the amount (in moles) of Magnesium oxide that oneelement will make:(.486 g O) / (16.0 g O) × (2 moles MgO)= .0608 moles MgO(.783 g Mg) / (24.3 g Mg) = .0322 moles MgOThere is less MgO produced with magnesium than oxygen; therefore, magnesium is the limiting reactant and the oxygen is the excess reactant. The magnesium determines how much Magnesium oxide is produced. It would be good to get .0608 moles of MgO, but there isn't enough magnesium. So the amount of MgO produced will be determined on the amount of Magnesium.Convert moles of MgO produced with the amount of oxygen to grams:.0322 mol MgO (40.3 g) = 1.30 grams of MgO produced--------------------------------------------------------------------------------------------------------You will need 3 moles of oxygen if you start with six moles of magnesium. This will allow you to produce 6 moles of magnesium oxide.Source: (e2020)
This is an incomplete question for several reasons. Iron bromide can be FeBr2 or FeBr3 and this will influence the answer. Also, in order to precipitate silver bromide, one would add silver nitrate, but it isn't stated how much silver nitrate is used. Assuming FeBr2 is used, and there is excess AgNO3, then .... 2AgNO3 + FeBr2 ----> 2AgBr(s) + Fe(NO3) moles FeBr2 = 2.96 g x 1 mol/216 g = 0.0137 moles moles AgBr produced = 2 x 0.0137 = 0.0274 moles mass AgBr produced = 0.0274 moles x 188 g/mole = 5.15 grams <---answer
1 mole of substance contains 6.02 x 1023 of its constituent particles. Accordingly, there would be 5.85 moles (of the substance) in a sample containing 3.52 x 1024 particles.
1 mole of any gas at STP occupies 22.4 liters. Thus, 2 moles propane will occupy 2 x 22.4 L = 44.8 liters.
The chemical formula for Calcium Carbide is CaC2. So to find how many moles are in 0.244 grams of CaC2, you must first find the molar mass. Ca = 40.08 g/mol C = 12.01 g/mol Now that you have the molar mass, you just add it all together. Since there are two Carbon, you add it's mass twice. 40.08 + 12.01 + 12.01 = 64.1g Now you use Stoichiometry to convert the .244g to moles. 0.244g * (1 mol) / (64.1 g) = .00380655 mol The final answer would round to 0.00381 mol
Glucose? C6H12O6 + 6O2 -> 6CO2 + 6H2O 6 moles water from one mole sugar.
If the reaction were 2AlCl3 + 3Pb ---> 3PbCl2 + 2Al, then from this balanced equation, 14 moles of AlCl3 would produce 3/2 x 14 moles of PbCl2 = 21 moles of PbCl2.
First you need to find the balanced reaction:2S + 3O2 --> 2SO3So using the balanced reaction we see that for every 3 moles of oxygen consumed, 2 moles of sulfur trioxide are produced:1.2 moles O2 consumed * (2 moles SO3/3 moles O2) = 0.8 mole of SO3 produced
The reaction would be H2 + 3N2 ==>2NH3moles H2 used = 5.69104 g x 1 mole/2.00 = 2.84552 moles H2moles NH3 produced (assuming N2 is NOT limiting) = 2 moles NH3/mole H2 x 2.84552 moles H2 = 5.69104 moles NH3 producedMolecules of NH3 produced = 5.69104 moles x 6.02x10^23 molecules/mole = 3.4x10^24 molecules
2 KClO3 ----> 2KCl + 3O2 So 2 moles of Potassium Chlorate produces 3 moles of oxygen molecules or 6 moles of oxygen atoms. 3 moles of Potassium chlorate would thus produce 4.5 moles of oxygen molecules or 9 moles of oxygen atoms.
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Since both chloride anions and nitrate anions have a charge of -1, there will be the same number of moles of silver chloride produced as the moles of silver nitrate reacted. (Since both silver nitrate and silver chloride are ionic compounds, it would be preferable to call their "moles" "formula units" instead.)
Assuming the reaction is S + O2 --> SO2, this equation is balanced as written, with everything in a 1:1 molar ratio. So, 67.1 moles of product would require 67.1 moles of O2 reactant.
It measures the amount of reactants actually produced in a reaction compared to the amount that would theoretically be produced if 100% of the reactants were converted to products according to the stoichiometry of the reaction. It is found by: actual moles of products ÷ predicted moles of products * 100%
4.8/16 moles of oxygen atoms converts to 1.6/16 moles of ozone molecules.
If you mean atoms then two, if molecules one.