Tungsten metal can be produced by the reaction of tungsten oxide powder (WO3) with hydrogen gas (H2) in a reactor heated to 800oC. The by-product of the reaction is water vapor (H2O). The reactor is charged with 128.0 kg of WO3 and 41.0 kg of H2. Assume that the reaction goes to completion.

Enter the number of kg of WO3 remaining:

This is a limiting reagent problem. Here is a post I've done explaining the steps for limiting reagent problems. Just follow the steps.

http://www.jiskha.com/science/chemistry/limiting_reagent_problem.html

To find the number of kilograms of WO3 remaining after the reaction, we'll need to determine the stoichiometric ratio between WO3 and H2. By looking at the balanced chemical equation for the reaction, we can see that the ratio is 1:3. This means that for every 1 mole of WO3, we need 3 moles of H2 to react completely.

First, we'll convert the masses of WO3 and H2 to moles using their respective molar masses. The molar mass of WO3 is 231.84 g/mol, and the molar mass of H2 is 2.02 g/mol.

Mass of WO3 = 128.0 kg = 128,000 g
Number of moles of WO3 = Mass of WO3 / Molar mass of WO3 = 128,000 g / 231.84 g/mol = 552.076 mol

Mass of H2 = 41.0 kg = 41,000 g
Number of moles of H2 = Mass of H2 / Molar mass of H2 = 41,000 g / 2.02 g/mol = 20,297.03 mol

Since the stoichiometric ratio is 1:3, every 1 mole of WO3 will react with 3 moles of H2. Therefore, the maximum moles of WO3 that can react completely is 552.076 mol.

To find the remaining moles of WO3, we'll subtract the moles of WO3 that reacted from the initial moles of WO3.

Remaining moles of WO3 = Initial moles of WO3 - Moles of WO3 that reacted
= 552.076 mol - 552.076 mol
= 0 mol

Finally, we'll convert the remaining moles of WO3 back to mass:

Remaining mass of WO3 = Remaining moles of WO3 * Molar mass of WO3
= 0 mol * 231.84 g/mol
= 0 g (or 0 kg)

Therefore, the number of kilograms of WO3 remaining after the reaction is 0 kg.