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.
28 grams of Nitrogen is necessary to produce 34 grams of ammonia.
Nitrogen is the limiting reactant and 4.15g of ammonia are produced.
there would be 50 grams of ammonia will be formed
16.0 g
How many grams of nitrogen dioxide must react with water to produce 5.00 x 1022 molecules of nitrogen monoxide?
28 grams of Nitrogen is necessary to produce 34 grams of ammonia.
Nitrogen is the limiting reactant and 4.15g of ammonia are produced.
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.
16,45 g nitrogen are needed.
there would be 50 grams of ammonia will be formed
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
394.794 grams
2g
16.0 g
The gram molecular mass of ammonia is 17.03. The formula shows that only one atom of nitrogen is required for each mole of ammonia; 4.12 mol of diatomic nitrogen contains 8.24 mol of nitrogen atoms, and with excess H, all of this nitrogen can be converted to ammonia. Therefore, 8.24 mol of ammonia can be produced, and multiplying this number by17.03 yields a total mass of 140.3 grams of ammonia, to the justified number of significant digits.
How many grams of nitrogen dioxide must react with water to produce 5.00 x 1022 molecules of nitrogen monoxide?
The complete question: Lead (II) oxide reacts with ammonia forming solid lead nitrogen gas and liquid water. 1.)How many grams of ammonia are consumed in the reaction of 75.0g lead (II) oxide? 2.) If 56.4g of lead are produced how many grams of nitrogen are also formed?