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
The answer is 20,664 g ammonia.
89,6 g ammonia are obtained.
To determine how many grams of NH3 can be produced from 3.64 g of H2, we first need to use the balanced chemical equation for the synthesis of ammonia: N2 + 3H2 → 2NH3. The molar mass of H2 is approximately 2.02 g/mol, so 3.64 g of H2 is about 1.80 moles. According to the reaction, 3 moles of H2 produce 2 moles of NH3, so 1.80 moles of H2 would produce approximately 1.20 moles of NH3. The molar mass of NH3 is about 17.03 g/mol, resulting in about 20.4 grams of NH3 produced.
0,044 moles of NH3 can be produced.
To determine the mass of NH3 produced from 2.22 mol of N2, we use the balanced equation for the synthesis of ammonia: N2 + 3H2 → 2NH3. From the equation, 1 mole of N2 produces 2 moles of NH3. Therefore, 2.22 mol of N2 will yield 2 × 2.22 = 4.44 mol of NH3. The molar mass of NH3 is approximately 17.03 g/mol, so the mass produced is 4.44 mol × 17.03 g/mol = 75.7 grams of NH3.
The answer is 20,664 g ammonia.
The mass of ammonia will be 95,03 g.
89,6 g ammonia are obtained.
4.33 mol of N2 will produce twice as many moles of NH3 since the balanced chemical equation is N2 + 3H2 -> 2NH3. Therefore, 4.33 mol of N2 will produce 8.66 mol of NH3. To convert this to grams, use the molar mass of NH3 (17.03 g/mol) to find that 8.66 mol is equal to 147.43 grams of NH3.
0,044 moles of NH3 can be produced.
To determine the mass of NH3 produced from 2.22 mol of N2, we use the balanced equation for the synthesis of ammonia: N2 + 3H2 → 2NH3. From the equation, 1 mole of N2 produces 2 moles of NH3. Therefore, 2.22 mol of N2 will yield 2 × 2.22 = 4.44 mol of NH3. The molar mass of NH3 is approximately 17.03 g/mol, so the mass produced is 4.44 mol × 17.03 g/mol = 75.7 grams of NH3.
The balanced chemical equation for the reaction is: N2 + 3H2 -> 2NH3 From the equation, it can be seen that 1 mol of N2 produces 2 mol of NH3. Therefore, 3.38 mol of N2 will produce 2 x 3.38 = 6.76 mol of NH3. To convert this to grams, you need to multiply the molar mass of NH3 (17.03 g/mol) by the number of moles of NH3 produced. Thus, 6.76 mol of NH3 will produce 6.76 x 17.03 = 115.18 g of NH3.
394.794 grams
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.
How many moles of NH3 are produced when 1.2 mol of nitrogen reacts with hydrogen?
How many moles of NH3 are produced when 1.2 mol of nitrogen reacts with hydrogen?
To find out how many grams of N2 are needed to produce 1.7 grams of NH3, you need to look at the balanced chemical equation for the reaction. For the reaction N2 + 3H2 -> 2NH3, the molar ratio of N2 to NH3 is 1:2. So you would need half the number of grams of N2 as NH3, which is 0.85 grams of N2.