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Calculation in two steps:
- mass (g) / molar mass (g/mol) = moles NH4NO3
6.84 / (1*14.01 + 4*1.008 + 1*14.01 + 3*16.00) =
6.84 / (80.052) = 0.0854 mole NH4NO3
- NH4NO3 moles * 2 moles N/mole NH4NO3 = moles N atoms
0.00854*2 = 0.171 moles N atoms
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Does one mole of NH3 weigh more than one mole of H2O?
The mass of NH3 mole = its molecular weight = 14 + 3 x 1 = 17 The mass of H2O mole = its molecular weight = 2 x 1 + 16 = 18 This means that one mole of NH3 weigh less than one mole of H2O
How does the experimental mole ratio compare with the theoretical mole ratio?
the experimental mole ratio has a bigger penis
How many grams of H2 are needed to produce 10.06 g of NH3?
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 the product to determine how many moles of product were yielded. According to the reaction, three moles of H2 are required to produces 2 moles of NH3. So, the mole ratio is 2/3. Multiply the number of moles of H2 with the molar ratio to determine the moles of NH3. 4.0606 * 2/3 = 3.3071 moles of NH3. Multiply the number of moles with the molar mass of NH3 (17.0306), and voila! 3.3071 * 17.0306 = 56.3216 grams. Now, if your teacher is feeling like a stickler about significant figures, than that value should be rounded to 56 grams of NH3.
How many moles of nh3 are produced from 1.5 moles h2?
1 mole
What is the limiting reagent when 150.0 g of nitrogen react with 32.1 g of hydrogen?
assuming you are talking about the synthesis of ammonia (NH3) N2 + 3 H2 --> 2 NH3 150.0 g / 28g/mole = 5.357 moles of nitrogen 32.1 g / 2.0g/mole = 16.05 mole of H2 5.357 moles N2 x 2 mol NH3 / 1 mol N2 = 10.714 mole ammonia (theoretical) 16.05 moles H2 x 2 mol NH3 / 3 mol H2 = 10.7 mole ammonia (theoretical) Almost the same, but 10.7 is slightly smaller. The smallest theoretical was found using hydrogen so hydrogen is the limiting reactant.
What is the mole ratio of nitrogen to ammonia in the equation 1N2 3H2 equals NH3?
The balanced equation for the reaction is N2 + 3H2 --> 2NH3 Thus, the mole ratio of nitrogen to ammonia in the balanced equation is 1:2.
How many moles of nitrate oxide can be made from the reaction of 3.80 mole NH3 with 5.15 mole o2?
4NH3 + 5O2 -> 4NO + 6H2O I suspect NH3 limits. Let's see. 5.15 O2 ( 4 mole NH3/5 mole O2) = 4.12 mole NH3 you do not have that much ammonia, so it limits and drives the reaction. 3.80 mole NH3 (4 mole NO/4 mole NH3) = 3.80 moles of NO made
How many grams of NH3 can be produced from 4.46 mol of N2 and excess H2.?
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.
Does one mole of NH3 weigh more than one mole of H2O?
The mass of NH3 mole = its molecular weight = 14 + 3 x 1 = 17 The mass of H2O mole = its molecular weight = 2 x 1 + 16 = 18 This means that one mole of NH3 weigh less than one mole of H2O
How does the experimental mole ratio compare with the theoretical mole ratio?
the experimental mole ratio has a bigger penis
How many grams of NH3 can be produced from 12.11 g H2?
N2(g) + 3H2-> 2NH3(g) This is the balanced equation Note the mole ratio between N2, H2 and NH3. It is 1 : 3 : 2 This will be important. moles N2 present = 28.0 g N2 x 1 mole N2/28 g = 1 mole N2 present moles H2 present = 25.0 g H2 x 1 mole H2/2 g = 12.5 moles H2 present Based on mole ratio, N2 is limiting in this situation because there is more than enough H2 but not enough N2. moles NH3 that can be produced = 1 mole N2 x 2 moles NH3/mole N2 = 2 moles NH3 can be produced grams of NH3 that can be produced = 2 moles NH3 x 17 g/mole = 34 grams of NH3 can be produced NOTE: The key to this problem is recognizing that N2 is limiting, and therefore limits how much NH3 can be produced.
How many grams of hydrogen are there in 150 g of ammonia?
150 grams NH3 (1 mole NH3/17.034 grams)(3 mole H/1 mole NH3)(1.008 grams/1 mole H)= 26.6 grams hydrogen=================17 g of ammonia has 3 g of hydrogen.So 150 g of ammonia will have 26.5 g of hydrogen
Find out the mass of HCL needed to react completely with 0.20 g of ammonia gass?
Use grams to moles to moles to grams: 0.2 g of ammonia gas (NH3) is equivalent to 0.012 moles of NH3 (divide by 17g/mole) One mole of NH3 reacts with one mole of HCl: NH3 + HCl <=> NH4Cl So we need 0.012 moles of HCl to react with 0.012 moles of NH3 0.012 moles HCl * 36.5 g/mole HCl => 0.43 g HCl
How do you calculate the mass in grams of 4.2 x 1025 NH3?
Molecules of ammonia? Will assume so. 4.2 X 1025 molecules NH3 (1 mole NH3/6.022 X 1023)(17.034 grams/1 mole NH3) = 1188 grams of ammonia ===================( could call it 1200 grams NH3 for significant figure correctness )
How many atoms of hydrogen are in 2NH3?
1 mole NH3 (3 mole H/1 mole NH3) = 3 mole hydrogen atoms
How many moles of hydrogen are required to produce 18.00 moles of ammonia?
N2 + 3H2 -> 2NH3 The stoichiometric equation (or balanced equation) for the formation of ammonia from this we can read off the mole ratio between hydrogen and ammonia; 3M H2 needed to produce 2M NH3 times each by 9 (so the ratio remains the same and 18M NH3 is formed) 27M H2 needed to produce 18M NH3
How many moles are in 7.95E22 molecules of NH3?
7.95 X 1022 molecules NH3 (1 mole NH3/6.022 X 1023) = 0.132 moles ammonia =================
