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25 grams / (17 grams/mole) x 6.022x1023 molecules/mole = 8.9x1023 molecules
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∙ 14y ago
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∙ 9y agoThere are no molecules.There are four atoms.
Wiki User
∙ 12y agoAnswer: 4.3 gram
Each mole of NH3 is 14 g + 3 times 1 g = 17 g in total,
so:
0.25 mole = 0.25 x 17 = 4.3 gram
Wiki User
∙ 10y ago1g of NH3 is 0.5871645822324moles (divide by 17.031). Now multiply this by avagadros number of 6.022x10^23, and you get 3.536x10^23 molecules.
Wiki User
∙ 8y agoNH3 = 17g/mol, and 25/17 = about 1.5mol. 1.5 x (6x10^23) = 9x10^23, which represents the number of NH3 molecules. NH3 = 4 atoms per molecule, so (9x10^23) x 4 = 3.6x10^24 atoms total.
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∙ 8y ago25 grams of nitrogen trihydride (NH3) is equivalent to 1,468 moles.
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How many molecules are in 75g of N2O3?
1 mole N2O3 = 76.0116g N2O3 (atomic weights x subscripts) 1 mole N2O3 = 6.022 x 1023 molecules N2O3 75g N2O3 x (1mol N2O3/76.0116g N2O3) = 0.99mol N2O3 0.99mol N2O3 x (6.022 x 1023 molecules N2O3/1mol N2O3) = 6.0 x 1023 molecules N2O3
How many molecules (not moles) of NH3 are produced from 5.95104 g of H2?
The number of ammonia molecules is 59 720.10e23.
Results for Calculate The Amount In Moles Of Molecules For Each Substance. A) A Sample Of Ammonia NH3(g) Containing 8.1x1020 Atoms Of Hydrogen?
NH3 Molecules = ( 8.1 x 10^20 H atoms ) ( 1 NH3 molecule / 3 H atoms ) NH3 Molecules = 2.7 x 10^20 NH3 molecules NH3 moles = ( NH3 molecules ) / ( N Avogadro ) NH3 moles = ( 2.7 x 10^20 NH3 molecules ) / ( 6.022 x 10^23 molecules / mole ) NH3 moles = 4.48 x 10^-4 NH3 moles <--------------
How many molecules -not moles- of NH3 are produced from 3.07104 g of H2?
Balanced equation is N2 + 3H2 ==> 2NH33.07104 g H2 x 1 mol/1.0079 g = 1.7679 moles H2 presentmoles NH3 produced = 2/3 x 1.7679 moles = 1.1786 moles NH3 formedmolecules NH3 = 1.1786 moles x 6.022x10^23 molecules/mole = 7.098x10^23 molecules (4 sig figs based on sig figs used in 6.022x10^23)
How many molecules (not moles) of NH3 are produced from 5.69104 g of H2?
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
How many molecules (not moles) of NH3 are produced from 1.57104 g of H2?
The answer is 1,57.10e27 molecules.
How many molecules are in 75g of N2O3?
1 mole N2O3 = 76.0116g N2O3 (atomic weights x subscripts) 1 mole N2O3 = 6.022 x 1023 molecules N2O3 75g N2O3 x (1mol N2O3/76.0116g N2O3) = 0.99mol N2O3 0.99mol N2O3 x (6.022 x 1023 molecules N2O3/1mol N2O3) = 6.0 x 1023 molecules N2O3
How many molecules (not moles) of NH3 are produced from 5.95104 g of H2?
The number of ammonia molecules is 59 720.10e23.
Results for Calculate The Amount In Moles Of Molecules For Each Substance. A) A Sample Of Ammonia NH3(g) Containing 8.1x1020 Atoms Of Hydrogen?
NH3 Molecules = ( 8.1 x 10^20 H atoms ) ( 1 NH3 molecule / 3 H atoms ) NH3 Molecules = 2.7 x 10^20 NH3 molecules NH3 moles = ( NH3 molecules ) / ( N Avogadro ) NH3 moles = ( 2.7 x 10^20 NH3 molecules ) / ( 6.022 x 10^23 molecules / mole ) NH3 moles = 4.48 x 10^-4 NH3 moles <--------------
How many grams are their in 3.4x10molecules of NH3?
The mass is 9,6.10e-22 g for 34 molecules.
How many molecules -not moles- of NH3 are produced from 3.07104 g of H2?
Balanced equation is N2 + 3H2 ==> 2NH33.07104 g H2 x 1 mol/1.0079 g = 1.7679 moles H2 presentmoles NH3 produced = 2/3 x 1.7679 moles = 1.1786 moles NH3 formedmolecules NH3 = 1.1786 moles x 6.022x10^23 molecules/mole = 7.098x10^23 molecules (4 sig figs based on sig figs used in 6.022x10^23)
How many molecules (not moles) of NH3 are produced from 5.69104 g of H2?
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
What is the mass of 1.75 1024 molecules of NH3?
74.168 u NH3Determine relative mass of NH3 molecule: (1 x 14.007 u N) + (3 x 1.008 u H) = 42.357 uMultiply the number of molecules of NH3 by its relative molecular mass.1.751024 molecules NH3 x (42.357 u NH3)/(1 molecule NH3) = 74.168 u NH3
How many grams does 7.74 x 1025 molecules of ammonia (NH3) represent?
This value is 21,85 g.
How many atoms are in 63.9g of ammonia?
Ammonia = NH3 and has a molar mass of 17.031 g/molmoles NH3 = 63.9 g x 1 mol/17.031 g = 3.752 molesmolecules NH3 = 3.752 moles x 6.02x10^23 molecules/mole = 2.259x10^24 moleculesEach molecule of NH3 has 4 atoms (1 N + 3 H), thus....number of atoms = 4 atoms/molecule x 2.259x10^24 molecules = 9.04x10^24 atoms (to 3 sig figs)
How many grams of NH3 can be produced from 2.08 grams of N2?
Ok, so I'm assuming that the chemical formula is written as - 3H2 + N2 ----> 2NH3 2.80 = moles of N2 17.03052 g/mol = Molar mass of NH3 (2.80 mol N2) x (2 NH3) / (1 N2) = 5.6 mol NH3 x (17.03052 g) / (1 mol NH3) = 95.4 g NH3
If you have 500 grams of NH3 how many moles to you have Please help and explain how you got the answer?
The molar mass of ammonia is 17,031 g.17,031 g NH3-------------------1 mol500 g NH3-------------------------xx= 500/17,031 = 29,36 moles
