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How many grams of ccl4 are needed to have 5.000 mol?
To find the grams of CCl4 needed, multiply the number of moles (5.000 mol) by the molar mass of CCl4 (153.82 g/mol). (5.000 , \text{mol} \times 153.82 , \text{g/mol} = 769.1 , \text{g}) Therefore, 769.1 grams of CCl4 are needed.
How many grams of NH3 will have the same number of molecules at 15.0 g of C6H6?
You first need to find the mass weight of NH3. wt. of N + (wt. of H)= 14.0067 + 3(1.0067)= 17.03052 Now that you have the mass weight, you divide 15 into 17.03052... 15.0/17.03052 = 0.8807 moles in 15. g of NH3
How many grams of fructose are in 1.20 moles of fructose?
The molar mass of fructose is approximately 180.16 g/mol. To find the mass in grams, you would multiply the number of moles (1.20 mol) by the molar mass (180.16 g/mol). Therefore, 1.20 moles of fructose would be 216.19 grams.
What is the mass in grams of 1.00 mol Al?
7 X 102 g Al this is the awnser using significant figures
How many grams in argon?
The molar mass of argon is approximately 40 g/mol. Therefore, one mole of argon weighs around 40 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 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 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 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 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.
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.
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
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.
How many grams of NH3 can be produced from 2.63 mol of N2 and excess H2?
89,6 g ammonia are obtained.
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.
How many grams of H2 are needed to produce 12.58 grams of NH3?
12.01 g NH3 = 0.667 mol NH3 = 1 mol H2 = 2.0 g H2
The formula for ammonia is NH3 the mass of 3 moles of ammonia would be how many grams?
The molar mass of ammonia is about 17 grams, so that 3 moles would have a mass of 51 grams.
