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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.
How many grams of NH3 can be produced from 4.33 mol of N2 and excess H2. Express your answer numerically in grams.?
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.
How many grams of NH3 can be produced from 2.08 grams of N2?
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
How many grams of NH3 can be produced from 2.23 mol of N2 and excess H2.?
By balancing the chemical equation for the reaction N2 + 3H2 -> 2NH3, we can see that 1 mol of N2 produces 2 mol of NH3. Therefore, 2.23 mol of N2 will produce 2.23 x 2 = 4.46 mol of NH3. Since the molar mass of NH3 is approximately 17 g/mol, 4.46 mol of NH3 is equivalent to 4.46 x 17 = 75.82 grams of NH3.
How many grams of NH3 can be produced from 3.38 mol of N2 and excess H2.?
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.
N2 plus 3H2 -- 2NH3 If you produce 55.5 grams of ammonia how many grams of nitrogen will you need?
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 NH3n N2 required = ( 3.259 mol NH3 ) ( 1 N2 mol / 2 NH3 mol ) = 1.629 moles N2m N2 required = ( 1.629 mol N2 ) ( 28.103 g N2 / mol N2 ) = 45.67 g N2 needed
What is the mass of 10 moles of ammonia given that molecular weight is 17?
If ammonia (NH3) has a molar weight of 17 grams per mole, then 10 moles means 17x10=170 grams.
How many grams of NH3 can be produced from 2.22 mol of N2 and excess H2.?
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.
How many grams of rm NH 3 can be produced from 2.90 mol of rm N 2?
To calculate the amount of NH3 that can be produced, we need to use the balanced chemical equation for the reaction between N2 and H2 to form NH3. The balanced equation is: N2 + 3H2 -> 2NH3. From the equation, we can see that 1 mole of N2 produces 2 moles of NH3. Therefore, if 2.90 mol of N2 is used, it will produce 2.90 mol * (2 mol NH3/1 mol N2) = 5.80 mol of NH3. To convert this to grams, we need to multiply the number of moles of NH3 by its molar mass, which is 17 g/mol. Therefore, the amount of NH3 produced from 2.90 mol of N2 is 5.80 mol * 17 g/mol = 98.6 g of NH3.
What is the molarity of a solution that contains 17g of NH3 in 0.50l of solution?
Molarity = moles of solute/Liters of solution need to find moles NH3 16.7 grams NH3 (1 mole NH3/17.034 grams) = 0.9804 moles NH3 --------------------------------now Molarity = 0.9804 moles NH3/1.50 Liters = 0.654 M -------------
What is the molecular mass of ammonia NH3?
17 g/mol is the molecular mass of ammonia NH3.
How many grams of NH3 can be produced from 4.10 moles of N and excess H?
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 --> 2NH34.10 mol N2 x (2 mol NH3)/(1 mol N2) x (17.031 g NH3)/(1 mol NH3) = 140. g NH3 = 1.40 x 10^2 g NH3 rounded to three significant figures.
