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How many grams of oxygen are required to burn 100 gram of propane?
For every 1 mole of propane burned, 5 moles of oxygen are required. This means that 44 grams of propane requires 160 grams of oxygen to burn completely. Therefore, 100 grams of propane would require (100 grams propane * 160 grams oxygen / 44 grams propane) = 363.64 grams of oxygen to burn completely.
How many grams of Oxygen are needed to completely react with 5.0 grams of butane in a butane lighter?
To calculate the grams of oxygen needed, you first need to balance the chemical equation for the combustion of butane. C₄H₁₀ + O₂ → CO₂ + H₂O. From the balanced equation, 2 moles of butane react with 13 moles of oxygen. One mole of butane is 58.12 g, and one mole of oxygen is 32 g. Therefore, 5.0 g of butane would require (5.0 g / 58.12 g/mol) * 13 moles of oxygen, which is approximately 1.12 grams of oxygen.
How many grams of oxygen are need to react with 4.6 grams of titanium chloride?
To find the grams of oxygen needed, we first calculate the molar mass of titanium chloride (TiCl4) and oxygen (O2). Then, we use the molar ratio of TiCl4 to O2 from the balanced chemical equation to find the grams of O2 needed.
How many grams of oxygen are needed to react with 23.9 grams of ammonia?
The balanced equation for the reaction between ammonia (NH3) and oxygen (O2) is 4NH3 + 5O2 → 4NO + 6H2O. To find the grams of oxygen needed to react with 23.9 grams of ammonia, you need to calculate the molar ratio between ammonia and oxygen using the balanced equation. Once you find the molar ratio, you can calculate the grams of oxygen required.
How many grams of oxygen are needed to react with 350 grams of iron to produce 500 grams of fe203?
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.
What is the total mass of water formed when 8 grams of hydrogen reacts completely with 64 grams of ovygen?
The balanced chemical equation for the reaction of hydrogen and oxygen to form water is 2H2 + O2 -> 2H2O. Based on the equation, for every 2 grams of hydrogen, 64 grams of oxygen are needed to form 36 grams of water. Thus, if 8 grams of hydrogen react completely with 64 grams of oxygen, the total mass of water formed would be 36 grams.
How many grams of oxygen are required to burn 100 gram of propane?
For every 1 mole of propane burned, 5 moles of oxygen are required. This means that 44 grams of propane requires 160 grams of oxygen to burn completely. Therefore, 100 grams of propane would require (100 grams propane * 160 grams oxygen / 44 grams propane) = 363.64 grams of oxygen to burn completely.
what is the mass in grams of oxygen, is needed to complete combustion of 6 L of methane?
what is the mass in grams of oxygen, is needed to complete combustion of 6 L of methane?
How many grams of Oxygen are needed to completely react with 5.0 grams of butane in a butane lighter?
To calculate the grams of oxygen needed, you first need to balance the chemical equation for the combustion of butane. C₄H₁₀ + O₂ → CO₂ + H₂O. From the balanced equation, 2 moles of butane react with 13 moles of oxygen. One mole of butane is 58.12 g, and one mole of oxygen is 32 g. Therefore, 5.0 g of butane would require (5.0 g / 58.12 g/mol) * 13 moles of oxygen, which is approximately 1.12 grams of oxygen.
What mass of oxygen (O2) is required to react completely with 86.17 grams of C6H14?
The answer is 152 g oxygen.
How many grams of oxygen are need to react with 4.6 grams of titanium chloride?
To find the grams of oxygen needed, we first calculate the molar mass of titanium chloride (TiCl4) and oxygen (O2). Then, we use the molar ratio of TiCl4 to O2 from the balanced chemical equation to find the grams of O2 needed.
How many grams of oxygen are needed to react with 23.9 grams of ammonia?
The balanced equation for the reaction between ammonia (NH3) and oxygen (O2) is 4NH3 + 5O2 → 4NO + 6H2O. To find the grams of oxygen needed to react with 23.9 grams of ammonia, you need to calculate the molar ratio between ammonia and oxygen using the balanced equation. Once you find the molar ratio, you can calculate the grams of oxygen required.
How many grams of oxygen gas are required to react completely with 14.6 grams of solid sodium to form sodium oxide?
Balanced equation. 4Na + O2 ->2Na2O 14.6 grams Na (1 mole Na/22.99 grams)(1 mole O2/4 mole Na)(32.0 grams/1 mole O2) = 5.08 grams oxygen gas needed --------------------------------------------
How many grams of oxygen are needed to react with 350 grams of iron to produce 500 grams of fe203?
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.
How many grams of oxygen will be required to produce 25 moles of sulfur dioxide how many grams of oxygen will be required to produce 25 moles of sulfur dioxide?
800 g oxygen are needed.
How many grams of potassium iodide will produce 750 grams of silver iodide when there is an excess of silver nitrate?
530,3 g potassium iodide are needed.
How many grams of oxygen are needed to react with 6.78 grams of ammonia?
To calculate the number of grams of oxygen needed to react with 6.78 grams of ammonia, we first write out the balanced chemical equation for the reaction between ammonia (NH3) and oxygen (O2) to form nitrogen monoxide (NO) and water (H2O). Then we use the stoichiometry of the equation to find the molar ratio between ammonia and oxygen. Finally, we convert the mass of ammonia to moles and then use the molar ratio to find the mass of oxygen needed.
