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How many moles s are present in 1 g of ammonia?
1 g of ammonia (NH3) is equal to 0,059 mol.
How many moles of which reactant will remain if 1.39 moles of N and 3.44 moles of H will react to form ammonia find out how many grams of ammonia can be formed and how many moles of limiting reactant?
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.
How many grams are 4.5 moles of NH3?
Here is the solution,for 1mole NH3 or ammonia,N: 14.0067 g/mol x 1H: 1.00794 g/mol x 3So approximately it is--> 17.03052 g/mol)Now multiply this into 4.5 for 4.5 moles of ammonia, --> 4.5x17.03052 =76.64g
How many moles of NH3 can be produced from the reaction of 75g of N2?
0,044 moles of NH3 can be produced.
How many moles of H2 are needed to produce 12.36 of NH3?
You can make a simple balance. There are (12.36 * 3) moles of H You have 2*H to form H2. So take the total from ammonia and divide by two to find the moles of H2 required.
How many moles are present in 1 g of ammonia?
There are approximately 0.023 moles of ammonia in 1 g of ammonia (NH3).
How many moles are present in 1g of ammonia?
To calculate the number of moles in 1g of ammonia (NH3), you first need to determine the molar mass of ammonia. The molar mass of NH3 is approximately 17 g/mol. Then you can use the formula: number of moles = mass / molar mass. So for 1g of NH3, there would be approximately 0.059 moles present.
How many moles s are present in 1 g of ammonia?
1 g of ammonia (NH3) is equal to 0,059 mol.
How many molecules of ammonia NH3 are present in 0.522 moles of ammonia?
0,522 moles of ammonia contain 3,143.10e23 molecules of NH3.
How many moles of electrons are present in 17 grams of ammonia?
To find the number of moles of electrons in ammonia (NH3), we first need to calculate the number of moles of ammonia using its molar mass. The molar mass of NH3 is 17 g/mol. Therefore, 17 grams of NH3 is equal to 1 mole. Since there are 3 electrons in each molecule of ammonia, there are 6.022 x 10^23 electrons present in 1 mole of NH3.
How many moles of oxygen gas are needed to react with 10.0 mol of ammonia?
The balanced chemical equation for the reaction between ammonia (NH3) and oxygen gas (O2) is 4 NH3 + 5 O2 → 4 NO + 6 H2O. This means that 5 moles of O2 are needed to react with 4 moles of NH3. With 10.0 moles of NH3, you would need 12.5 moles of O2 (10.0 moles NH3 x 5 moles O2 / 4 moles NH3).
How many moles of hydrogen atoms are contained in 3 moles of ammonia?
Since ammonia has a chemical formula of NH3, it contains one mole of nitrogen and three moles of hydrogen per mole of ammonia. Therefore, 3 moles of ammonia contain 3 moles of nitrogen and 9 moles of hydrogen atoms.
How many moles in 100grams of ammonia?
The molecular formula of ammonia is NH3The molecular mass of NH3 is 14.0 + 3(1.0) = 17.0 Amount of NH3 present = mass of sample/molar mass = 100/17.0 = 5.88g
100 grams of ammonia equals how many moles?
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
How many moles of ammonia is 170000grams?
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.
How many moles of which reactant will remain if 1.39 moles of N and 3.44 moles of H will react to form ammonia find out how many grams of ammonia can be formed and how many moles of limiting reactant?
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.
How many moles are in 12 x 10.3 grams of ammonia NH3?
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.
