How many moles of NH3 can be formed from 3.0 mole N2 and 6.0 mole H2?
N2+3 H2----- 2 NH3 First of all we have to find the limiting reactant 1 mole of N2 requires=3 moles of H2 2 moles of N2 require=6moles of H2 However there are 3 moles of N2 for 6 moles of H2 So N2 is in excess H2 is our limiting reactant. 3 moles of H2 make =2 moles of NH3 1 mole of H2 makes=2/3 moles of NH3 6 moles of H2=2/3 *6. moles of NH3 Moles of NH3=4 moles Hope it helps :)
If 2 moles of H2 react with O2, how many moles of H2O are formed?
If [math]{H_{2} + \frac {1}{2} O_{2} -> H_{2}O}[/math], then [math]{2H_{2} + O_{2} -> 2 H_{2}O}[/math]
How many moles of nh3 can be formed when .50 mol of NCl3 is allowed to react with .40 mol of h2?
2 NCl3 + 3 H2 = 2 NH3 + 3 Cl2 if 0.5 moles of NCl3 react we need 0.50 x 3 / 2 moles of H2 = 0.75 moles but we only have 0.40 moles of H2 so it is the limiting reagent and we get 0.40 x 2 / 3 moles of NH3 = 0.267 moles of NH3 produced by these moles of H2
What will be the volume of NH3 formed when 2 moles of N2 reacts with 6 moles of H2?
Do your own homework.
How many moles of NH3 can be formed from 3.0 moles of N2 and 6 moles of H2?
First you write the equation for formation of ammonia from hydrogen and nitrogen:N2 + 3H2 → 2NH3Note that nitrogen and hydrogen react in the ratio 1:3. The reactants that you have are in the ratio 1:2. Thus you have more nitrogen than you need to react with the hydrogen that you have - nitrogen is in excess (or, you don’t have enough hydrogen to react with all your nitrogen). You then write a balanced equation using the actual moles of the reagent that isn’t in excess (hydrogen):2N2 + 6H2 → 4NH3And you can see that you will form 4 moles of ammonia. You will also have one mole of nitrogen left over as you only needed 2 moles to react with your hydrogen.
1.) If 12.0 mol of H2(g) react with excess N2(g), how many moles of ammonia are produced?
1.) If 12.0 mol of H2(g) react with excess N2(g), how many moles of ammonia are produced? 2.) Equation for question---Sn(s)+2HF(g)--->SnF(s)+H2(g) If 45.0 grams of HF react with Sn, how many grams of stannous fluoride are produced? 3.) Equation for question---CH4(g)+2O2(g)--->C02(g)+2H2O(... If 150 moles of carbon dioxide are produced, what mass, in grams, of methane is required? 4.) Equation for question---4AI(s)+3O(g)--->2A12O3(s) what mass, in grams, of the product would be produced if 625 mL of oxygen react at STP?
How many moles of NH3 are produced when 1.8 moles H2 reacts?
3 H2 + N2 → 2 NH3 (1.8 mol H2) x (2 mol NH3 / 3 mol H2) = 1.2 mol NH3
How many moles of H2 are needed to react with 5.3 mol of O2?
It will take twice the amount (10.6 moles) to react with that amount of molecular oxygen due to the 2-to-1 mol ratio for these two reactants, assuming the product is water.
How many moles of NH3 can be made if 15 moles of H2 react with enough N2?
To answer the question you first need to determine what the balanced equation is for the reaction. The reaction is3H2 + N2 → 2NH3So, for every 3 moles of H2 you get 2 of ammonia (NH3). So multiply the amount of hydrogen by the ratio 2/3 and you will get the answer. Thus2/3 * 15 =10 moles of NH3.
How many moles of NH3 can be produced from 15.0mol of H2 and excess N2?
How many moles of NH3 can be produced from 15.0mol of H2 and excess N2? Not many, since the reaction does not go to completion. But that's a different story. We'll pretend that the reaction does go to completion... 3H2(g) + N2(g) --> 2NH3(g) 15.0mol ............... ? mol 15.0 mol H2 x (2 mol NH3 / 3 mol H2) = 10.0 mol NH3 This is the unit-factor method, the preferred way for solving stoichiometry problems. In the unit-factor method you use one or more conversion factors. You set the problem up to cancel out the units you don't want and leave the units you do. In your case the conversion factor is (2 mol NH3 / 3 mol H2) which comes from the balanced chemical equation. "Moles H2" cancel out, leaving "moles NH3". Then it is simply a matter of doing the corresponding math.