The mass is 9,6.10e-22 g for 34 molecules.
394.794 grams
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.
For this you need the atomic (molecular) mass of NH3. Take the number of moles and multiply it by the atomic mass. Divide by one mole for units to cancel. NH3=16.0 grams100 grams NH3 / (16.0 grams) = 6.25 moles NH3
The answer is 20,664 g ammonia.
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.
Molecules of ammonia? Will assume so. 4.2 X 1025 molecules NH3 (1 mole NH3/6.022 X 1023)(17.034 grams/1 mole NH3) = 1188 grams of ammonia ===================( could call it 1200 grams NH3 for significant figure correctness )
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
25 grams / (17 grams/mole) x 6.022x1023 molecules/mole = 8.9x1023 molecules
To find the grams of H2 needed, we first calculate the moles of NH3 using its molar mass. Then, we use the balanced chemical equation to determine the mole ratio of H2 to NH3. Finally, we convert moles of H2 to grams using its molar mass.
To find the number of molecules in 0.95 grams of SF6, you would first calculate the number of moles of SF6 using its molar mass. Then, using Avogadro's number, you can determine the number of molecules. Next, you would convert this number of molecules to the number of moles of NH3 using the mole ratio between NH3 and SF6. Finally, you would convert the moles of NH3 to grams using the molar mass of NH3 to find the mass needed.
16,45 g nitrogen are needed.
The mass of ammonia will be 95,03 g.