If you have 6 moles of N2 how many moles of NH3 can you make?
If you have 6 moles of N2 you can make theoretically 12 moles of NH3.
If 5.0 moles of NH3 are produced 2.5 moles of N2 are used.
What mass of NH3 is produced when 1.20 mol of N2 react completely in the following equation N2 plus 3H2 2NH3?
N2 + 3H2 ==> 2NH3moles N2 = 1.20 moles moles NH3 formed = 1.20 moles N2 x 2 moles NH3/1 moles N2 = 2.40 moles NH3 mass NH3 = 2.40 moles x 17 g/mole = 40.8 g NH3
I assume you mean this reaction. N2 + 3H2 ->2NH3 0.60 moles NH3 (1 mole N2/2 mole NH3) = 0.30 mole N2 reacts ================
Balanced equation: 3H2 + N2 ==> 2NH34.16 moles N2 x 2 mole NH3/1 mole N2 = 8.32 moles NH3 8.32 moles NH3 x 17 g/mole = 141 grams NH3
How many grams of NH3 can be produced from the reaction of 17.8 moles of H2 and a sufficient supply of N2 N2 plus 3 H2 and rarr 2 NH3?
N2 + 3H2 --> 2NH3 ... balanced equation moles H2 used = 17.8 moles of NH3 theoretically produced = 17.8 moles H2 x 2 moles NH3/3 moles H2 = 11.9 moles NH3
0,044 moles of NH3 can be produced.
How do you find if you have 14.5 moles of N2 how many moles of H2 are theoretically needed to produce 22.5 moles of NH3 according to the reaction N2 plus 3H2 2NH3?
N2 + 3H2 -> 2NH3 If you have moles produced you can do it this way. 22.5 moles NH3 (3 moles H2/2 moles NH3) = 33.8 moles H2 needed -----------------------------------
The answer is 0,2 moles nitrogen.
2 x 0.60 = 1.2 the reaction is N2 + 3H2 -> 2NH3 (1 mole of nitrogen N2 give 2 moles of NH3)
The answer is 42,55 g.
You need the balanced chemical equation for N2 and H2 combining to form ammonia, NH3. N2 (g) + 3 H2 (g) -----> 2 NH3 (g) Moles NH3 = ( 55.5 g NH3 ) / ( 17.03 g/mol NH3 ) = 3.259 moles of NH3 n N2 required = ( 3.259 mol NH3 ) ( 1 N2 mol / 2 NH3 mol ) = 1.629 moles N2 m N2 required = ( 1.629 mol N2 )… Read More
Write the balanced equation: N2 + 3H2 ==> 2NH3 7.9 moles N2 x 2 moles NH3/1 mole N2 = 15.8 moles NH3 formed (assuming H2 isn't limiting). Corrected for 2 significant figures, answer would be 16 moles
Ok, so I'm assuming that the chemical formula is written as - 3H2 + N2 ----> 2NH3 2.80 = moles of N2 17.03052 g/mol = Molar mass of NH3 (2.80 mol N2) x (2 NH3) / (1 N2) = 5.6 mol NH3 x (17.03052 g) / (1 mol NH3) = 95.4 g NH3
3H2+N2 --->2NH3 So you can double of moles N2.You can get 9.12mol NH3.
The volume is 4908 L at oC.
Balanced equation. N2 + 3H2 --> 2NH3 1.4 moles H2 (2 moles NH3/3 moles H2) = 0.93 moles NH3 produced =======================
The molecular mass of NH3 is the sum of the atomic mass of nitrogen and three times the atomic mass of hydrogen, or 14.007 + 3(1.008) = 17.031. Therefore, the number of moles of NH3 in 14.0 grams is 14.007/17.031 = 0.822. Since each molecule of N2 supplies two nitrogen atoms and each molecule of NH3 needs only one nitrogen atom, the number of moles of N2 needed is half the number of moles of… Read More
The Synthesis of Ammonia is done through Haber's Process: N2 + 3H2 -> 2NH3 so 2 moles of NH3 requires 3 moles of H2 gas so if X moles of NH3 requires 19.5 moles of H2 gas then X=(2/3)*19.5=13 NH3 formed = 13 moles
3H2 + N2 ----> 2NH3, right? 2.01 = moles of N2 17.03052 g/mol = Molar mass of NH3 (2.01 mol N2) x (2 NH3) / (1 N2) = 5.6 mol NH3 x (17.03052 g) / (1 mol NH3) = 68.463 g NH3
To form NH3, I assume. Balanced equation. N2 + 3H2 -> 2NH3 0.85 moles N2 (3 moles H2/1 mole N2) = 2.6 moles hydrogen gas needed ---------------------------------------------
Lets assume the reaction: N2 + H2 -> NH3 Now to balance it: 1 N2 + 3 H2 -> 2 NH3 There is a 2/1 ratio for moles of NH3/N2 So 2 mol NH3 produced for every 1 mol of N2 reacted. 0.50 mol NH3 produced * (1 mol of N2 reacted / 2 mol NH3 produced) = 0.25 mol of N2 reacted.
3H2 + N2 <------> 2NH3 is the balanced equation for Hydrogen and Nitrogen making ammonia. 3 moles of H2 produces two moles of ammonia and thus to make 6 moles requires 9 moles of Hydrogen.
0,3 moles of nitrogen reacted.
N2+3H2 --->2NH3 3mol of H2 give 2mol NH3. So 16.5mol of H2 give 11mol of NH3
How many moles of ammonia gas are produced by the reaction of 3 mole of nitrogen gas with excess hydrogen gas N2 H2 ---- NH3?
N2 + 3H2 --> 2NH3 You have been told, indirectly, that nitrogen limits and will drive the reaction. 3 moles N2 (2 moles NH3/1 mole N2) = 6 moles ammonia gas produced ========================
The maximum amount ammonia that can be produced is 17 g. This is a limiting reagent (reactant) question. You will need to determine the mass of NH3 produced by each reactant. Start with a balanced equation. N2(g) + 3H2(g) --> 2NH3(g) You need to determine the molar masses of each gas. I'm going to round to whole numbers. N2: 2 x 14 g/mol N = 28 g/mol N2 H2: 2 x 1 g/mol H =… Read More
0.6 moles of N2 produce 1.2 moles of NH3
33 moles or nitrogen and 9 moles of hydrogen are needed.
How many moles of NH3 can be produced by the reaction of 2.00 g of N2 with 3.00 g H2 Reaction N2(g) plus 3 H2(g) 2 NH3(g)?
0,143 moles of NH3 are produced.
The molecular weight of NH3 is 17.03-grams per mole and 14.01 for N2. The reaction is N2 + 3H2 = NH3. Therefore for every 1-mole of N2 as a reactant 1-mole of NH3 is produced. .2941-moles of NH3 is produced with a mass of 5.01-grams.
0,325 moles of nitrogen are involved in the reaction.
There is no compound NH. However, there is ammonia, NH3. The reactants are nitrogen gas, N2, and hydrogen gas, H2. Multiply moles N2 by the mole ratio from the balanced equation between NH3 and N2, so that NH3 is in the numerator. Then multiply by the molar mass of NH3, 17.031 g/mol. Balanced equation: N2 + 3H2 --> 2NH3 4.10 mol N2 x (2 mol NH3)/(1 mol N2) x (17.031 g NH3)/(1 mol NH3) =… Read More
The mass of ammonia will be 95,03 g.
How many grams of NH3 are produced in the reaction with 23.1 moles of N2 in N2 plus 3H2 yields 2NH3?
The result is 786,8 g (theoretically).
First, balance the equation to get your proper molar ratios: N2 + 3H2 --> 2NH3 Now, plug in 7.5 moles of H2 instead of the coefficient of 3. A direct proportion may help: 3/2=7.5/x, solve for x to get 5 moles of NH3.
What volume of ammonia gas NH3 in liters is produced at STP by the complete reaction of 100 g of nitrogen N2 with excess hydrogen?
First lets convert the 100g N2 into moles (100/28=3.57 moles). Now lets look at the balanced reaction equation: N2 + 3H2 -> 2NH3 This means that if I have 3.57 moles of N2, I will make 7.14 moles of NH3. Now, assuming that everything can be modelled as an ideal gas: V=(nRT)/P V=(7.14*.0821*273)/1= 160.03 L
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.
The balanced equation for the formation of NH3 is N2 + 3 H2 --> 2 NH3. 13.64 grams of ammonia is equal to .801 moles. Then 1.2015 moles of hydrogen are needed, or 2.42 grams.
Steps to get the right answer 1. Look at the equation 2. The question is asking for how many moles of N2 reacted! 3. So that means you should look to start with 0.75moleNH3 4. You would want the NH3 to cancel so you look back at the equation and find that 2 is in front of NH3 to balance the equation. 5. What you should have now: [0.75mole NH3/2molesNH3] 6.Then you look back at… Read More
9,62 moles ammonia = 163.84 g
N2 + 3H2 ---> 2NH3 so 3 moles of hydrogen produces 2 moles of ammonia. Therefore with 6 moles of hydrogen available, 4 moles of ammonia only are possible as the hydrogen is the yield limiting material.
The reaction would be H2 + 3N2 ==>2NH3 moles H2 used = 5.69104 g x 1 mole/2.00 = 2.84552 moles H2 moles NH3 produced (assuming N2 is NOT limiting) = 2 moles NH3/mole H2 x 2.84552 moles H2 = 5.69104 moles NH3 produced Molecules of NH3 produced = 5.69104 moles x 6.02x10^23 molecules/mole = 3.4x10^24 molecules
Need balanced equation here. N2 + 3H2 -> 2NH3 1.5 moles H2 (2 moles NH3/3 moles H2) = 1.0 moles ammonia produced =======================
N2 + 3H2 ---> 2NH3 so 3 moles of Hydrogen produces 2 moles of ammonia. Thus 1.8 moles will produce 1.8/3 x 2 = 1.2 moles of ammonia.
The reaction between nitrogen and hydrogen to form ammonia is: N2 + 3 H2 â†’ 2 NH3 The above is the reaction for the Haber process in the industrial synthesis of ammonia. For a given proportion of 3 N2 to 2 H2 (or in ratio terms equivalent to 4.5 N2 to 3 H2), we see that H2 is the limiting reactant. Thus according to the stoichiometry of the reaction, 2 moles of H2 will form… Read More
I assume you are talking about the Haber Process: 3H2 + N2 ----> 2NH3 If 3.71 mol of N2 is produced, there will be 7.42 mol of ammonia produced (as per mol ratios). Using the formula, n = g/mw ---> g = 7.42 x 17.034 = 126.39228 grams of NH3
How many grams of NH3 can be produced from 2.08 grams of N2
moles of N2 = 1000/28= 35.7 moles moles of NH3 = 2 x 35.7= 71.4 moles mass of NH3 = 71.4 x 17 (14+1+1+1) = 1214g
Balanced equation is N2 + 3H2 ==> 2NH3 3.07104 g H2 x 1 mol/1.0079 g = 1.7679 moles H2 present moles NH3 produced = 2/3 x 1.7679 moles = 1.1786 moles NH3 formed molecules NH3 = 1.1786 moles x 6.022x10^23 molecules/mole = 7.098x10^23 molecules (4 sig figs based on sig figs used in 6.022x10^23)
The formula reaction for NH3 when using N2 and H2 is: (N2)+3(H2) ---> 2(NH3) Now, first step is to find the moles of the H2 reactant. This is found via (grams of reactant)/(molar mass of reactant). There are 10 grams, and the molar mass of H2 is approximately 2.016. Therefore, the equation should look like: 10/2.016. This yields a value of ~4.9606 moles of H2. Now, you use the molar ratio from the reactant to… Read More