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How many grams of oxygen are needed to react with 6.78 grams of ammonia?
To calculate the number of grams of oxygen needed to react with 6.78 grams of ammonia, we first write out the balanced chemical equation for the reaction between ammonia (NH3) and oxygen (O2) to form nitrogen monoxide (NO) and water (H2O). Then we use the stoichiometry of the equation to find the molar ratio between ammonia and oxygen. Finally, we convert the mass of ammonia to moles and then use the molar ratio to find the mass of oxygen needed.
If you start with seventy grams of nitrogen and twenty grams of hydrogen how many grams of ammonia will be formed and how many grams of hydrogen are leftover?
To form ammonia (NH3) from nitrogen (N2) and hydrogen (H2), the balanced chemical equation is N2 + 3H2 → 2NH3. This means that for every mole of nitrogen, 3 moles of hydrogen are required. Given that nitrogen is limiting in this case, all 70 grams of nitrogen will react with 210 grams (3 times 70) of hydrogen to form 70 grams of ammonia. This reaction will consume all the hydrogen, leaving no grams of hydrogen leftover.
What is the limiting reactant when 3.14g of nitrogen react with 2.79g of hydrogen to produce ammonia and how many grams of ammonia are produce?
To determine the limiting reactant, we need to find the moles of each reactant. Then, we calculate the amount of ammonia that can be produced from each reactant. Whichever reactant produces the least amount of ammonia is the limiting reactant. Finally, we calculate the grams of ammonia produced based on the limiting reactant.
The mass of 3 moles of ammonia would be how many grams?
3 moles of ammonia is 51grams. One mole is 17 grams.
How many moles are in 1200 grams 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.
How many grams of Pb form when 0.105 g NH3 react?
None. A reaction of ammonia does not produce any lead!
How many grams of oxygen are needed to react with 6.78 grams of ammonia?
To calculate the number of grams of oxygen needed to react with 6.78 grams of ammonia, we first write out the balanced chemical equation for the reaction between ammonia (NH3) and oxygen (O2) to form nitrogen monoxide (NO) and water (H2O). Then we use the stoichiometry of the equation to find the molar ratio between ammonia and oxygen. Finally, we convert the mass of ammonia to moles and then use the molar ratio to find the mass of oxygen needed.
How many grams of oxygen are needed to react with 23.9 grams of ammonia?
The balanced equation for the reaction between ammonia (NH3) and oxygen (O2) is 4NH3 + 5O2 → 4NO + 6H2O. To find the grams of oxygen needed to react with 23.9 grams of ammonia, you need to calculate the molar ratio between ammonia and oxygen using the balanced equation. Once you find the molar ratio, you can calculate the grams of oxygen required.
If you start with seventy grams of nitrogen and twenty grams of hydrogen how many grams of ammonia will be formed and how many grams of hydrogen are leftover?
To form ammonia (NH3) from nitrogen (N2) and hydrogen (H2), the balanced chemical equation is N2 + 3H2 → 2NH3. This means that for every mole of nitrogen, 3 moles of hydrogen are required. Given that nitrogen is limiting in this case, all 70 grams of nitrogen will react with 210 grams (3 times 70) of hydrogen to form 70 grams of ammonia. This reaction will consume all the hydrogen, leaving no grams of hydrogen leftover.
How many grams are there in 200 g of ammonia?
There are 200 grams of ammonia in 200 grams 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 grams of ammonia NH3 will produce 300 grams of N2?
394.794 grams
What is the limiting reactant when 3.14g of nitrogen react with 2.79g of hydrogen to produce ammonia and how many grams of ammonia are produce?
To determine the limiting reactant, we need to find the moles of each reactant. Then, we calculate the amount of ammonia that can be produced from each reactant. Whichever reactant produces the least amount of ammonia is the limiting reactant. Finally, we calculate the grams of ammonia produced based on the limiting reactant.
How many grams of ammonia NH3 will be prepared when 6.00g of hydrogen H2 has reacted?
The reaction between hydrogen and ammonia to form ammonia is 3H2 + N2 → 2NH3. To find the amount of ammonia produced when 6.00g of hydrogen reacts, first convert the mass of hydrogen to moles using its molar mass. Then, use the mole ratio from the balanced equation to find the moles of ammonia produced, and finally, convert this to grams using the molar mass of ammonia.
How many hydrogen molecules are needed to produce 525 grams 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.
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.
The mass of 3 moles of ammonia would be how many grams?
3 moles of ammonia is 51grams. One mole is 17 grams.
