For this you need the atomic (molecular) mass of N2. Take the number of moles and multiply it by the Atomic Mass. Divide by one mole for units to cancel.
2.25 moles × 28.0 grams = 63.0 grams N2
To calculate the mass of H2 needed to produce 13.14g of NH3, you can use the molar ratio between H2 and NH3. From the balanced chemical equation for the reaction, you can see that 2 moles of NH3 are produced for every 3 moles of H2 consumed. First, calculate the number of moles of NH3 using its molar mass, then use the molar ratio to find the moles of H2 needed, and finally convert the moles of H2 into grams.
Assuming you are talking about production from the Haber process, and also assuming Hydrogen is not the limiting quantity and also assuming 100% conversion, then the answer is as follows: N2 + 3H2 ----> 2NH3 Then 1 mole of N2 produces 2 moles of ammonia. So 3.94 moles of N2 produces 7.92 moles of ammonia. Taking the molecular weight of ammonia to be 17 then 17 x 7.92 = 134.64 g of ammonia produced.
The formula for the sum of the series r(1/n2-1/n2) is r(1-1/n2).
The only information you've given is you want to know moles and how many Liters there are. To calculate, you need to know what you are preparing(i.e., N2, AgNO3, etc.), and it's Molarity(M). Here's an example: How many moles of silver nitrate are needed to prepare 250mL of standard 0.100M silver nitrate solution? Note: M = moles of solute(stuff) / volume of solution(L) = # moles / L So, ?/250mL=.1M First we use dimensional analysis to convert mL to L. 250mL x 10-3 L / 1 mL = .25L Since we know M=moles/L, we can take the Molarity and put it in moles/L form. .10M = .10 mol/L Finally, .10 mol/L x .25L = .025 mol AgNO3 We can make a general assumption for your question and say the M is standard .100M. .10M = .10 mol/L .10 mol/L x 48L = 4.8 moles
The freezing point of a substance with a molecular weight of N2 is -210.01 degrees Celsius.
The answer is 0,0043 moles of N2.
The mass of ammonia will be 95,03 g.
In the reaction 3H2 + N2 --> 2NH3, the ratio of H2 to N2 is 3:1. To calculate the amount of N2 required, we need to first convert the mass of H2 to moles, then use the ratio to find the moles of N2 needed, and finally convert the moles of N2 to grams. After the calculation, we find that 2.79 g of H2 requires 3.31 g of N2 to react completely.
For this you need the atomic (molecular) mass of N2. Take the number of moles and multiply it by the atomic mass. Divide by one mole for units to cancel..713 moles × (69.7 grams) = 49.7 grams Ga
The molecular mass of NH3 is the sum of the atomic mass of nitrogen and three times the atomic mass of hydrogen, or 14.007 + 3(1.008) = 17.031. Therefore, the number of moles of NH3 in 14.0 grams is 14.007/17.031 = 0.822. Since each molecule of N2 supplies two nitrogen atoms and each molecule of NH3 needs only one nitrogen atom, the number of moles of N2 needed is half the number of moles of NH3 formed = 0.411.
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
To find the number of molecules of N2 in 3.5 grams, first calculate the number of moles using the molar mass of N2 (28 g/mol). Then use Avogadro's number (6.022 x 10^23 molecules/mol) to convert the moles to molecules.
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
One look at a periodic table tells you the molar mass of sodium (Na) is 22.99 g/mol. So multiply your mol value by the g/mol value to get a value in grams! 2.0x10-3 x 22.99 = !!!
If 5.0 moles of NH3 are produced 2.5 moles of N2 are used.
1 mole N2 = 28.0134g 1 mole N2 = 6.022 x 1023 molecules N2 28.0134g N2 = 6.022 x 1023 molecules N2 (4.00 x 1023 molecules N2) x (28.0134g/6.022 x 1023 molecules) = 18.6g N2
3,44 moles H2 react with 1,146 moles NH3. The limiting reactant is hydrogen. O,244 moles N2 remain. 19,5 g NH3 are obtained.