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Ba(NO3)2

The total mass for this molecule is 261g

so what is 261g/132g this is 1.97moles

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How many moles of oxygen atoms are contained in one mole of BaNO32?

9 moles (there are four oxygen atoms for every mol of BaSO4, so you multiply 2.25 by 4)


What is a procedure for making 100 ml of a 0.10 M solution of barium nitrate?

Ba(NO3)2 Molarity = moles of solute/Liters of solution ( 100 ml = 0.100 liters ) 0.10 M Ba(NO3)2 = moles Ba(NO3)2/0.100 liters = 0.01 moles Ba(NO3)2 (261.32 grams/1 mole Ba(NO3)2) = 2.6 grams of Ba(NO3)2 needs to be put into that 100 milliliters of solution.


How many oxygen atoms does Ba(NO3)2 contain?

Ba(NO3)2 contains a total of 6 oxygen atoms - there are 3 oxygen atoms in each nitrate (NO3) group, and there are 2 nitrate groups in Ba(NO3)2.


How many moles of BaCl2 barium chloride is necessary to react with the 7.5 moles AgNo3 silver nitrate in 2AgNO3 BaCl-- 2AgCl Ba(NO3)2?

3,75 moles barium chloride


How many moles ofBa3(PO4) could be formed from 3.25 moles of Na3PO4 and 7.03 moles of Ba(NO3(2 in the following reaction?

Ba3(PO4) is an incorrect formula. Barium phosphate has the formula 'Ba3(PO4)2 '. So write down the BALANCED reaction eq'n. 2Na3PO4 + 3Ba(NO3)2 = 6NaNO3 + Ba3(PO4)2 The molar ratios are 2:3 :: 6:1 We have 3.25 mol(Na3PO4) equivalent to '2' & 7.03 mol(Ba(NO3)2 ) equivalent to '3' We now must find the limiting reactant. So 3..25 / 2 = 1.625 & 7.03/3 = 2.34333.... We only need 1.625 moles(Na3PO4) to react with 7.03 moles (Ba(NO3)2). So at 3.25 moles the sodium phosphate is in excess. Therefore 7.03 moles barium nitrate is the limiting reactant. By equivalence 3:1 ;; 7.03 : x Hence x /1 = 7.03 / 3 x = 7.03 x 1 / 3 x = 2.34333... moles(Ba(NO3)2) is produced.


What is the mass of one mole of barium nitrate?

First, we need to find the formula for barium nitrate. Barium's charge is 2+, and nitrate is a 1-. If you criss-cross charges, the resulting formula is Ba(NO3)2. Now that we have the formula, we need to find the atomic mass. Barium has an atomic mass of 137. Nitrate is composed of nitrogen (14.01g) and oxygen (16.00*3 = 48). The atomic mass of nitrate is 62.01. Since there are two NO3 molecules, we multiply by 2 to get 124.02. Finally, we add the barium for a grand total of 261.02. Finally, we set-up our conversion ratios: 432 g Ba(NO3)2/1 * 1 mol/261.02g = 1.66 mols


Ba(NO3)2 contains how many oxygen atoms?

One mole of barium nitrate contains six moles of oxygen atoms. One formula contains 6 atoms.


Formula for barium nitrite?

The chemical formula for Barium Nitrite is Ba(NO2)2. Ba = Barium N = Nitrogen O = Oxygen Its molar mass is 229.338 g/mol.


Results for precipitation of Ba(NO3) and Na2SO4?

Ba(NO3)2(aq) + Na2SO4(aq) ==> BaSO4(s) + 2NaNO3(aq).


Formula for Ba2 plus and NO3-?

The chemical formula for Barium ion (Ba2+) is simply Ba2+ and the chemical formula for Nitrate ion (NO3-) is NO3−. To represent the compound formed by Barium ion and Nitrate ion, write Ba(NO3)2.


What is the number of moles in 471.6gBa(OH)2?

471,6 g of anhydrous Ba(OH)2 is equivalent to 2,75 moles.


What is Ba(NO3)2?

Ba(NO3)2 is the chemical formula for barium nitrate, a compound composed of one barium ion (Ba^2+) and two nitrate ions (NO3^-). It is commonly used in fireworks to produce a green flame.