There are approximately 8.91 mols in 18.0g of hydrogen gas.
Calculation: 18.0g x (1mol/2.02g) = 8.91 mols
One mole of hydrogen gas (H2) is 2.02g.
5 moles
0,0011 moles of hydrogen are obtained.
9H2 + 3N2 ------> 6NH3 3 moles of N2 would be required.
The balanced chemical equation for the reaction between oxygen and hydrogen is2H2 + 02 -> 2H2OThus 2.2 moles of oxygen reacts with 4.4 moles of hydrogen to form 4.4 moles of steam (water in gaseous state).The mass of H2O obtained is thus 4.4 x 18.0 = 79.2g.
How many hydrogen atoms are in 35.0 grams of hydrogen gas?
5 moles
For each mole of hydrogen gas (H2) reacting with chlorine gas (Cl2), you will get 2 moles of HCl. H2 + Cl2 = 2 HCl
Mass (g) = Mr * Moles If you rearrange it, you get Moles = Mass/Mr Working with a 2dp periodic table you get: Moles = 2/1.01 =1.98 There are 1.98 moles of hydrogen in 2g of H2 gas.
0,0011 moles of hydrogen are obtained.
750 L hydrogen gas at 0 0C and 1 at is equal to 33,44 moles.
Ideal gas equation. PV = nRT ===============
9H2 + 3N2 ------> 6NH3 3 moles of N2 would be required.
Since hydrogen is a gas, we would need more information to answer it. As chance wrote, you will need twice as much hydrogen as oxygen. However, in order to know what the volume of that hydrogen is, we also need to know the temperature and pressure so that we can use the universal gas law to get the answer.
Absolutely none, as there is no oxygen in hydrogen cyanide. Its formula is HCN--one atom each of hydrogen, carbon and nitrogen.
CO2 + H2 -> CO + H2O one to one here 30.6 moles H2O (1 mole H2/1 mole H2O) = 30.6 moles Hydrogen gas needed
N2 + 3H2 -> 2NH3 3 moles hydrogen gas. You should know that because of the formula of ammonia.
The balanced chemical equation for the reaction between oxygen and hydrogen is2H2 + 02 -> 2H2OThus 2.2 moles of oxygen reacts with 4.4 moles of hydrogen to form 4.4 moles of steam (water in gaseous state).The mass of H2O obtained is thus 4.4 x 18.0 = 79.2g.