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The equation for the formation of ammonia from nitrogen and hydrogen gases at standard temperature and pressure is N2 + 3 H2 -> 2 NH3. The gram molecular mass of ammonia is 17.03 and the gram molecular mass of divalent nitrogen is 28.0134. Therefore, designating the unknown mass of nitrogen needed as p, the proportion 660/p equals (2 X 17.03)/28.0134 is valid. This expression is algebraically equivalent to 660 = (34.06)p/28.0134, for which p = 543 grams, rounded to the justified number of significant digits, limited by the three significant digits specified by the given number 600.
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What mass of nitrogen is needed to make 34 grams of ammonia?
To find the mass of nitrogen needed to make ammonia, first determine the molar mass of ammonia (NH3) which is 17 g/mol. Since there is one nitrogen atom in ammonia, the nitrogen mass is 14 g/mol. To make 34 grams of ammonia, you would need 14 grams of nitrogen.
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 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.
How many grams of nitrogen dioxide must react with water to produce 5.00 x 1022 molecules of nitrogen monoxide?
To find the grams of nitrogen dioxide needed, first calculate the moles of nitrogen monoxide using Avogadro's number. Then, use the balanced chemical equation to determine the moles of nitrogen dioxide required. Finally, convert moles to grams using the molar mass of nitrogen dioxide.
What mass of nitrogen is needed to make 34 grams of ammonia?
To find the mass of nitrogen needed to make ammonia, first determine the molar mass of ammonia (NH3) which is 17 g/mol. Since there is one nitrogen atom in ammonia, the nitrogen mass is 14 g/mol. To make 34 grams of ammonia, you would need 14 grams of nitrogen.
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 are needed to produce 3.5 mole of nitrogen?
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.
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.
Stoichiometry grams of N2 are needed to produce 20 grams NH3 N2 plus H2 NH3?
16,45 g nitrogen are needed.
Ammonia gas NH3 can be manufactured by combining hydrogen and nitrogen gases If 28 g of nitrogen and 7 g of hydrogen are available how many grams of ammonia can be produced?
The balanced chemical equation for the formation of ammonia from nitrogen and hydrogen is N2 + 3H2 → 2NH3. From the equation, it can be seen that 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia. Calculate the moles of nitrogen and hydrogen provided, determine the limiting reactant, and then use stoichiometry to find the grams of ammonia that can be produced.
How much nitrogen is in 7.5 grams of ammonia?
First you need to find the atomic masses of each element involved in the compound NH3, and add them up to find the total molecular mass of ammonia.Nitrogen = 14.0 gramsHydrogen = 1.01 × 3 atoms = 3.03 grams----------------------------------------------------Ammonia = 17.03 gramsThen you take the mass of nitrogen in one molecule and divide it by the total mass to find the percent composition.14.0 grams Nitrogen ÷ 17.03 grams Ammonia = .822 = 82.2% nitrogen in ammoniaThen you simply need to take 82.2% of 7.5 grams to find how much nitrogen is in that particular amount.82.2% × 7.50 = 6.17 grams of nitrogen in 7.50 grams of ammonia
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.
Is ammonia vapor heavier than nitrogen?
Yes, ammonia vapor is heavier than nitrogen. The molecular weight of ammonia (NH3) is 17 grams per mole, while nitrogen (N2) has a molecular weight of 28 grams per mole. This difference in molecular weight causes ammonia vapor to be denser and heavier than nitrogen gas.
How many grams of ammonia NH3 will produce 300 grams of N2?
394.794 grams
How many grams of nitrogen dioxide must react with water to produce 5.00 x 1022 molecules of nitrogen monoxide?
To find the grams of nitrogen dioxide needed, first calculate the moles of nitrogen monoxide using Avogadro's number. Then, use the balanced chemical equation to determine the moles of nitrogen dioxide required. Finally, convert moles to grams using the molar mass of nitrogen dioxide.
