P = 31 mass weight
H = 1 mass weight, times that by 3 to get 3 (you do this because you have PH3)
Add both mass weights to get 34, then multiply that by the number of mols you have
34 x 3.91 mol = 130 g
The molar mass of oxygen (O) is approximately 16.00 grams/mol. Therefore, for 2.0 moles of oxygen, the mass would be 2.0 moles x 16.00 g/mol = 32.00 grams.
15.994 grams/mol, but it can rounded to 15.99 or 16.0 g/mol.
The least molecular mass of a compound containing 8% sulfur by mass would be the mass of sulfur (32 g/mol) divided by the mass percentage of sulfur (8%). This would give a molecular mass of 400 g/mol.
To find the mass of iron in CuFeS2, we first calculate the molar mass of CuFeS2: Cu (copper) has a molar mass of 63.5 g/mol, Fe (iron) has a molar mass of 55.8 g/mol, and S (sulfur) has a molar mass of 32 g/mol. Adding these together gives a molar mass of 151.3 g/mol for CuFeS2. Next, calculate the mass percentage of iron in CuFeS2 by dividing the molar mass of iron by the molar mass of CuFeS2 and multiplying by 100. The percentage of iron is approximately 37%, so in 86.6 grams of CuFeS2, the mass of iron would be 0.37 * 86.6 grams = 32 grams.
256.52 g mol-1since sulfur molecules are S8.
The balanced chemical reaction is: 4NH3(g) + 5O2(g) = 4NO(g) + 6H2O We use the reaction and the amount of the reactant , NH3, to determine the amount of oxygen needed. We do as follows: 200 g NH3 ( 1 mol / 17.04 g) (5 mol O2 / 4 mol NH3) (32 g / 1 mol) = 469.48 g O2 needed
Oxygen (1,429 g/cm3) is more dense than ammonia (o,70 g/cm3).
To find the mass of a 32 mol gold coin, you would first need to know the molar mass of gold, which is approximately 197 g/mol. Then, you can calculate the mass by multiplying the number of moles by the molar mass: 32 mol * 197 g/mol = 6304 g.
The mass of 0.75 mol of sulfur can be calculated by multiplying the number of moles by the molar mass. Therefore, the mass would be 0.75 mol x 32 g/mol = 24 grams.
The balanced chemical equation for the reaction between ammonia (NH3) and oxygen (O2) is 4NH3 + 3O2 → 2N2 + 6H2O. From the equation, we can see that 3 moles of O2 are needed to react with 4 moles of NH3. This means the molar ratio of O2 to NH3 is 3:4. First, calculate the number of moles of NH3 in 200.0 g: 200.0 g NH3 / 17.03 g/mol NH3 = 11.75 moles NH3 Now, calculate the number of moles of O2 needed using the molar ratio: 11.75 moles NH3 * (3 moles O2 / 4 moles NH3) = 8.81 moles O2 Finally, convert moles of O2 to grams: 8.81 moles O2 * 32 g/mol O2 = 282.0 g O2.
15.99 x 2 g
The molar mass of SO2 is 64.06 g/mol. Oxygen contributes 32 g/mol to this molar mass, so the mass percent of oxygen in SO2 is (32 g/mol / 64.06 g/mol) x 100, which is approximately 50%.
To calculate the amount of NH3 needed to react with 21 grams of CH3OH, you first need to balance the chemical equation for the reaction. Then, you convert the mass of CH3OH to moles and use the mole ratio from the balanced equation to determine the moles of NH3 required. Finally, convert the moles of NH3 to grams using the molar mass of NH3.
The formula mass of CaSO4 is calculated by adding the atomic masses of each element in the compound. Ca has an atomic mass of 40 g/mol, S has an atomic mass of 32 g/mol, and O has an atomic mass of 16 g/mol. So, the formula mass of CaSO4 is 40 + 32 + (4 x 16) = 136 g/mol.
Assuming you mean the decomposition of ammonia: 2NH3 --> N2 + 3H2, 100g NH3 = 5.88mol (100/17), and the ratio of ammonia to nitrogen is 2:1. 5.88mol/2=2.94mol, so that's the amount of nitrogen produced. 2.94x28 (molar mass of N2) gets you 82.3g nitrogen produced.
The molar mass of O2 is 32 g/mol, H2 is 2 g/mol, and N2 is 28 g/mol. To calculate the mass in grams, you would multiply the molar mass by the number of moles of the substance. For example, for 1 mole of O2, the mass would be 32 grams.
One mole of sulfur atoms has a mass of approximately 32 grams. This is based on the molar mass of sulfur, which is 32 g/mol.