Calculation in two steps:
6.84 / (1*14.01 + 4*1.008 + 1*14.01 + 3*16.00) =
6.84 / (80.052) = 0.0854 mole NH4NO3
0.00854*2 = 0.171 moles N atoms
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
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.
The experimental mole ratio is obtained from actual experimental data, while the theoretical mole ratio is calculated based on the balanced chemical equation. Comparing the two can reveal discrepancies and provide insight into the accuracy of the experimental results or any potential sources of error in the experiment.
The balanced chemical equation for the reaction is: 3H2 + N2 -> 2NH3 From the stoichiometry of the balanced equation, 3 moles of H2 produces 2 moles of NH3. Therefore, if you have 1.5 moles of H2, you will produce 1.5 moles * (2 moles NH3 / 3 moles H2) = 1 mole of NH3.
N2 + 3H2 ==> 2NH3moles N2 = 1.20 molesmoles NH3 formed = 1.20 moles N2 x 2 moles NH3/1 moles N2 = 2.40 moles NH3mass NH3 = 2.40 moles x 17 g/mole = 40.8 g NH3
The mole ratio of nitrogen to ammonia in the equation is 1:2. This means that for every 1 mole of nitrogen (N2) that reacts, 2 moles of ammonia (NH3) are produced.
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
First, calculate the number of moles of H2 using its molar mass of 2 g/mol. Then, use the balanced chemical equation for the reaction between H2 and NH3 to determine the mole ratio. Finally, convert the moles of H2 to moles of NH3 using the mole ratio and then calculate the number of molecules of NH3 produced.
nitrogen weighs 14, hydrogen weighs 1, so NH3 weighs 14+(3x1)=17grams
To find the number of ammonia molecules in 3.75g of CH4, you first need to convert 3.75g of CH4 to moles using the molar mass. Then, you need to use the mole ratio between CH4 and NH3 to find the number of ammonia molecules in the given amount of CH4. The mole ratio for CH4 to NH3 is 1:3, as 1 molecule of CH4 produces 3 molecules of NH3 in the balanced chemical equation.
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
For the reaction N2 + 3H2 -> 2NH3, the mole ratio between N2 and NH3 is 1:2. Therefore, if 0.40 mole of NH3 is produced, 0.20 moles of N2 reacted.
The molecular mass of ammonia (NH3) is 18.03 grams/mole
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.
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
1 mol of N2 reacts with 3 mol of H2 to produce 2 mol of NH3. Therefore, 4.46 mol of N2 will produce (4.46 mol x 2 mol NH3 / 1 mol N2) = 8.92 mol of NH3. As NH3 has a molar mass of 17.03 g/mol, 8.92 mol of NH3 is equal to (8.92 mol x 17.03 g/mol) = 152 g of NH3.
1:3 because...in the equation N2+3H2---> 2NH3the product is 2NH3... there are 2 Nitrogens and 6 Hydrogens (2*3) making the ratio 2:6; but the correct simplified answer is 1:3