To determine how many grams of Mercury (Hg) are needed to produce 3.75 moles of oxygen (O₂), we first consider the balanced chemical reaction for the formation of oxygen from mercury(II) oxide (HgO): 2 HgO → 2 Hg + O₂. From this, we see that 2 moles of HgO produce 1 mole of O₂, which means 3.75 moles of O₂ require 7.5 moles of Hg. The molar mass of Hg is approximately 200.59 g/mol, so 7.5 moles of Hg would weigh about 1,504.43 grams (7.5 moles × 200.59 g/mol).
To calculate the moles of hydrogen needed to produce 68 grams of ammonia (NH₃), we start with the balanced chemical equation for the synthesis of ammonia: N₂ + 3H₂ → 2NH₃. The molar mass of ammonia is approximately 17 g/mol, so 68 grams of NH₃ corresponds to 68 g / 17 g/mol = 4 moles of NH₃. Since 3 moles of hydrogen are required for every 2 moles of ammonia, the moles of hydrogen needed is (4 moles NH₃) × (3 moles H₂ / 2 moles NH₃) = 6 moles of H₂. Therefore, 6 moles of hydrogen must react to produce 68 grams of ammonia.
800 g oxygen are needed.
The answer is 699 moles perchloric acid.
These reagents doesn't react.
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.
When 2.5 moles of oxygen react with hydrogen, they react in a 1:2 ratio to produce water. Therefore, 2.5 moles of oxygen will produce 5 moles of water. To convert moles to grams, you'll need to know the molar mass of water, which is approximately 18 grams/mol. So, 2.5 moles of oxygen will produce 90 grams (5 moles x 18 grams/mole) of water.
The balanced chemical equation for the reaction between iron and oxygen to produce Fe2O3 is 4Fe + 3O2 -> 2Fe2O3. From the equation, we see that 3 moles of oxygen react with 4 moles of iron to produce 2 moles of Fe2O3. Therefore, to find the grams of oxygen needed, we need to calculate the molar mass of Fe2O3 and then determine the number of grams needed using the mole ratio from the balanced equation.
To calculate the moles of hydrogen needed to produce 68 grams of ammonia (NH₃), we start with the balanced chemical equation for the synthesis of ammonia: N₂ + 3H₂ → 2NH₃. The molar mass of ammonia is approximately 17 g/mol, so 68 grams of NH₃ corresponds to 68 g / 17 g/mol = 4 moles of NH₃. Since 3 moles of hydrogen are required for every 2 moles of ammonia, the moles of hydrogen needed is (4 moles NH₃) × (3 moles H₂ / 2 moles NH₃) = 6 moles of H₂. Therefore, 6 moles of hydrogen must react to produce 68 grams of ammonia.
To find the grams of nitrogen dioxide needed, first calculate the moles of nitrogen monoxide using Avogadro's number. Then, use the balanced chemical equation to determine the moles of nitrogen dioxide required. Finally, convert moles to grams using the molar mass of nitrogen dioxide.
1 mole of P4O10 reacts with 6 moles of water to produce 4 moles of H3PO4. Therefore, 10.0 moles of water will produce (10/6)*4 moles of H3PO4. To convert moles to grams, multiply the number of moles by the molar mass of H3PO4.
Given the balanced chemical equation: 4Al + 3O2 → 2Al2O3, we can see that 4 moles of aluminum react with 3 moles of oxygen to produce 2 moles of aluminum oxide. In this case, 18.32 grams of aluminum is equivalent to 0.684 moles. Using stoichiometry, we find that this would produce 0.456 grams of aluminum oxide.
800 g oxygen are needed.
The answer is 699 moles perchloric acid.
These reagents doesn't react.
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.
To produce 1 mole of chloroform, you need 3 moles of chlorine. So, to produce 1.5 moles of chloroform, you would need 4.5 moles of chlorine. Converting moles to grams by using the molar mass of chlorine (35.5 g/mol) gives you 160.5 grams of chlorine required.
The mass of diborane is 442,7 g.