3 moles of ammonia is 51grams. One mole is 17 grams.
To find the number of moles in 1200 grams of ammonia, divide the given mass by the molar mass of ammonia. The molar mass of ammonia (NH3) is approximately 17 grams/mole. Therefore, 1200 grams divided by 17 grams/mole equals approximately 70.59 moles of ammonia.
Ammonia = NH3Molecular mass = 16.0Formula of grams to moles: grams / molecular mass = moles170,000 g / (16.0) = 10,600 moles NH3Note that the answer is with three significant digits
For this you need the atomic (molecular) mass of NH3. Take the number of moles and multiply it by the atomic mass. Divide by one mole for units to cancel. NH3=16.0 grams100 grams NH3 / (16.0 grams) = 6.25 moles NH3
The reaction of nitrogen with hydrogen to form ammonia is: N2 +3H2 = 2NH3 Therefore to make 10 moles of ammonia you need 5 moles N2 and 15 moles H2
4.25 grams. .050 M = .050 mol/1 L 5.0 L x .050 mol/L (cancel out L to get mol as a unit)= .25 mol Atomic mass of Ammonia (NH3)= 17 g/mol .25 mol x 17 g/mol (cancel out mol to get g as a unit)= 4.25 g
The molar mass of ammonia is about 17 grams, so that 3 moles would have a mass of 51 grams.
To find the number of moles in 170000 grams of ammonia, you need to divide the given mass by the molar mass of ammonia. The molar mass of ammonia (NH3) is about 17 grams/mol. Therefore, 170000g ÷ 17g/mol ≈ 10000 moles of ammonia.
To find the number of moles in 1200 grams of ammonia, divide the given mass by the molar mass of ammonia. The molar mass of ammonia (NH3) is approximately 17 grams/mole. Therefore, 1200 grams divided by 17 grams/mole equals approximately 70.59 moles of ammonia.
To produce 525 grams of ammonia (NH3), you would need 25 moles of ammonia. Since the balanced chemical equation for the reaction between hydrogen and nitrogen to form ammonia is 3H2 + N2 -> 2NH3, you would need 75 moles of hydrogen molecules (H2) to produce 525 grams of ammonia. This is equivalent to 4,500 molecules of hydrogen.
That amount of ammonia contains two moles of hydrogen gas. One mole of hydrogen gas weighs 2.016 grams. Therfore 3.75 grams of ammonia contains two moles of hydrogen.
To calculate the moles of hydrogen needed to produce 68 grams of ammonia (NH₃), we start with the balanced chemical equation for the synthesis of ammonia: N₂ + 3H₂ → 2NH₃. The molar mass of ammonia is approximately 17 g/mol, so 68 grams of NH₃ corresponds to 68 g / 17 g/mol = 4 moles of NH₃. Since 3 moles of hydrogen are required for every 2 moles of ammonia, the moles of hydrogen needed is (4 moles NH₃) × (3 moles H₂ / 2 moles NH₃) = 6 moles of H₂. Therefore, 6 moles of hydrogen must react to produce 68 grams of ammonia.
Ammonia = NH3Molecular mass = 16.0Formula of grams to moles: grams / molecular mass = moles170,000 g / (16.0) = 10,600 moles NH3Note that the answer is with three significant digits
For this you need the atomic (molecular) mass of NH3. Take the number of moles and multiply it by the atomic mass. Divide by one mole for units to cancel. NH3=16.0 grams100 grams NH3 / (16.0 grams) = 6.25 moles NH3
The mass of ammonia will be 95,03 g.
this is a easy one. There are only 0.04166 moles.
To find the number of moles in 12 x 10.3 grams of ammonia (NH3), you first need to calculate the molar mass of NH3 (17.03 g/mol). Then divide the given mass (12 x 10.3 g) by the molar mass to get the number of moles, which should be approximately 72 moles.
3,44 moles H2 react with 1,146 moles NH3. The limiting reactant is hydrogen. O,244 moles N2 remain. 19,5 g NH3 are obtained.