Nitrogen and hydrogen gas react to form ammonia gas as follows: N2(g) + 3H2(g) -> 2NH3(g).?
The answer is 2.2L. Keep in mind that Avogadro's law says that a given volume of gas at the same temperature and pressure has the same number of moles of gas molecules, regardless of what gas that is. So 1L of H2 at a given temp and pressure has the same number of molecules as 1L of F2, NH3, N2, CH4, etc, at the same temperature and pressure. As a result of that, you can use the given volumes to calculate product yields instead of having to calculate how many moles you have. Since both react completely, neither here is the limiting reagent. So it doesn't matter whether you do the calculation with N2 or H2. Here's how it goes: 1.1L N2 x (2 NH3)/(1 N2) = 2.2L NH3 Alternatively: 3.3L H2 x (2 NH3)/(3 H2) = 2.2L NH3
Hydrogen gas reacts with nitrogen gas to form ammonia gas according to the equation?
Hydrogen gas reacts with nitrogen gas to form ammonia gas according to the equation 3H2(g) + N2(g) ---> 2NH3(g) 1. How many grams of ammonia can be produced from 1.62 mols of nitrogen? 2. How many grams of hydrogen gas are required to produce ammonia with a mass of 12.00 g ? 3. How many molecules of ammonia are produced from 7.72×10^−4 g of hydrogen?
Nitrogen (N2) and hydrogen (H2) react to form ammonia (NH3).?
NImra's answer is correct, but I can tell you HOW to get it! The balanced equation is: N2 + 3H2 --> 2 NH3 You are starting with 6 N2 molecules and 6 H2 molecules (and 0 NH3 molecules) This is actually an excess/limiting reagent problem. In order to completely react with all the N2 present ( 6 molecules), the molecules of H2 NEEDED is calculated using the coefficient ratio of H2 to N2 (3:1) N2 + 3H2 --> 2 NH3 : 6 molecules N2 x (3 molecules H2)/(1 molecule N2) = 18 molecules H2. Now you NEED 18 H2 molecules but you only HAVE 6 H2 molecules present, so H2 is the LIMITING REAGENT! Which means all 6 molecules of H2 will react and there will some N2 (the EXCESS REAGENT) left over. To calculate the molecules of NH3 produced use the coefficient ratio of NH3 to H2, which is 2:3, N2 + 3H2 --> 2 NH3 6 molecules H2 x (2 molecules NH3)/(3 molecules H2) = 4 molecules NH3 Now, use the 6 molecules of the Limiting Reagent (H2), to calculate the # of molecules of N2 which react. N2 + 3H2 --> 2 NH3 The coefficient ratio of N2 to H2 is 1:3. 6 molecules H2 x (1 molecule N2)/(3 molecules H2) = 2 molecules N2 have reacted. Since 6 molecules of N2 were initially present and 2 molecules reacted, (6 - 2 = 4 molecules of N2 are left over). So in the final mixture you have 4 molecules of N2, 0 molecules of H2 (the LR is completely consumed), and 4 molecules of NH3.
The stoichiometric equation is:3H2 + N2 → 2NH3According to the stoichiometric equation, 6g of H2 combines with 28g of N2.So, 1.09g of H2 will combine with = (1.09 × 28)/6 = 5.0866g of N2.Now, 14 of N2 is used, according to the question.So, H2 is the limiting reagent and N2 is the excess reagent.Now, according to the stoichiometric equation, 6g of H2 produces 34g of NH3.So, 1.09g of H2 will produce = (1.09 × 34)/6 = 6.176g of NH3.
Hydrogen gas, H2, reacts with nitrogen gas, N2, to form ammonia gas, NH3, according to the equation 3H2(g)+N2(g)→2NH3(g)?
Part A How many moles of NH3 can be produced from 21.0 mol of H2 and excess N2? Express your answer numerically in moles. Part B How many grams of NH3 can be produced from 3.77 mol of N2 and excess H2. Express your answer numerically in grams. Part C How many grams of H2 are needed to produce 13.63 g of NH3? Express your answer numerically in grams. Part D How many molecules (not moles) of NH3 are produced from 3.74×10−4 g of H2? Express your answer numerically as the number of molecules.
Nitrogen gas (N2) reacts with hydrogen gas (H2) to form ammonia (NH3) at 200°C in a closed container. 1.04 atm
Nitrogen gas (N2) reacts with hydrogen gas (H2) to form ammonia (NH3) at 200°C in a closed container. 1.04 atm of nitrogen gas is mixed with 2.24 atm of hydrogen gas. At equilibrium the total pressure is 2.24 atm. Calculate the partial pressure of hydrogen gas at equilibrium. can someone please explain this to me i have a test this week and this was one of the review questions
Hydrogen gas, H2, reacts with nitrogen gas, N2, to form ammonia gas, NH3.?
you need to write the balanced chemical reaction 3H2 + N2>>>> 2NH3 FROM THE REACTION YOU NEED TO USE THE STOICHMETRIC FACTOR PARENTHESIS MEAN MULTIPLICATION MOLECULAR WEIGHT FOR NH3= 42g/mol 2.96 MOLES OF N2(2 MOLES OF NH3/1 MOLE OF N2)(MW OF NH3/1 MOLES OF NH3=GRAMS OF NH3 THAT CAN BE PRODUCED. NOW USE THE SAME STRATEGY TO FIND THE GRAMS OF H2 NEEDED TO PRODUCE 14.70g of NH3 TO FIND THE MOLECULES YOU HAVE TO USE AVOGADRO'S NUMBER. 8.57×10^−4g of H2(1 MOLE OF H2/MW OF H2)(2 MOLES NH3/2 MOLES H2)(6.02*10^23 MOLECULES OF NH3/1 MOLE OF NH3)= MOLECULES OF NH3 REMEMBER TO BALANCE THE CHEMICAL EQ, USE THE STOICHMETRIC FACTOR TO GET THE QUANTITIES YOU NEED. REMEMBER THAT IN 1 MOLES OF THINGS YOU HAVE 6.02*10^23 THINGS.
Hydrogen gas, H2, reacts with nitrogen gas, N2, to form ammonia gas, NH 3, according to the equation 3H2 (g) + N2 (g) → 2NH3 (g)The reaction is reversible and the production of ammonia is exothermic.Some notes on the conditionsThe catalystThe catalyst is actually slightly more complicated than pure iron. It has potassium hydroxide added to it as a promoter - a substance that increases its efficiency.The pressureThe pressure varies from one manufacturing plant to another, but is always high. You can't go far wrong in an exam quoting 200 atmospheres.RecyclingAt each pass of the gases through the reactor, only about 15% of the nitrogen and hydrogen converts to ammonia. (This figure also varies from plant to plant.) By continual recycling of the unreacted nitrogen and hydrogen, the overall conversion is about 98%.
Limiting reagent in a reaction is the one which is consumed totally before the total consumption of any other reactant and for the calculation of the same, it is better to do the calculations on mole basis.So, for the given condition first let we calculate the number of moles of each reactant-As we know,Number of moles = Weight (in grams)/Molecular Wt.Mol. Wt. of N2- 28Mol. Wt. of H2- 2Since, we have 28g of N2 & 12g of H2Thus,No. of moles of N2- 28/28 = 1 mole& No. of moles of H2- 12/2 = 6 molesNow, from the stoichiometry of the reactionN2 + 3H2 → 2NH3We can see that, for 1 mole of N2 we require 3 moles of H2 to produce 2 moles of NH3 and here for 1 mole of N2 we have 6 moles of H2 i.e. we have 3 moles of excess H2 which will be unused. Thus for the given case N2 will be the limiting reagent.Thank you.P.S.- In stoichiometric problems, always do calculations on the molar basis.
OK, when ammonia (NH3) is made from hydrogen (H2) and nitrogen (N2) the balanced equation of the reaction is:3 H2 + N2 → 2 NH3and for purposes of doing this problem it is a little easier to write it such that you only get one NH3 rather than two so divide by two and you get3/2 H2 + 1/2 N2 → NH3Next, how much, in moles, is a ton of ammonia? First convert tons to grams:1 ton * 2000 lb/ton * 0.454 kg/lb * 1000 g/kg = 908,000 g.The molecular weight of ammonia is 17.031 g/mol so 908,000 grams is:908,000 g /17.031 g/mole = 53,315 molSo, to produce one ton of ammonia you will need:1.5 H2 * 53,315 mol = 79,971 mol H20.5 N2 * 53,315 mol = 26,657 mol N2If you want the numbers in grams you just multiple the amount of H2 in moles by the molecular weight of H2 which is 1.008 g/mol. Likewise, for N2 using the molecular weight of N2 which is 14.007 g/mol.