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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 required to react with 100 g Fe in this reaction?
The balanced chemical equation for the reaction is: 4Fe + 3O2 -> 2Fe2O3 From the equation, it can be seen that 3 moles of O2 are required to react with 4 moles of Fe. Therefore, to determine the grams of O2 required to react with 100 g Fe, you would need to use stoichiometry to find the answer.
How many grams of gas are needed to react with 348.5 grams of oxygen?
To determine the amount of gas needed to react with 348.5 grams of oxygen, you need to know the balanced chemical equation of the reaction. Then, use the stoichiometry of the reaction to calculate the amount of gas required based on the molar ratio between the gas and oxygen in the reaction.
What is the mole ratio of oxygen to pentane?
The mole ratio of oxygen to pentane in the balanced chemical equation for the combustion of pentane is 13:1. This means that 13 moles of oxygen are required to completely react with 1 mole of pentane.
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 fe2o3 are required to completely react with 84 grams of co?
160...cant quite grasp HOW though
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 much oxygen in grams will be required to react with 0.5 mole of aluminium?
Three atoms of oxygen are required to react with each two atoms of aluminum to form the most common product of reaction between oxygen and aluminum. Therefore, 0.75 mole of oxygen atoms will be required to react with 0.5 mole of aluminum atoms. The atomic weight of oxygen is 15.999; therefore, the mass will be (0.75)(15.999) = 12 grams of oxygen, to the maximum possibly justified number of significant digits.
How many grams of Oxygen are required to react with 100 g Fe in this reaction?
The balanced chemical equation for the reaction is: 4Fe + 3O2 -> 2Fe2O3 From the equation, it can be seen that 3 moles of O2 are required to react with 4 moles of Fe. Therefore, to determine the grams of O2 required to react with 100 g Fe, you would need to use stoichiometry to find the answer.
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 gas are needed to react with 348.5 grams of oxygen?
To determine the amount of gas needed to react with 348.5 grams of oxygen, you need to know the balanced chemical equation of the reaction. Then, use the stoichiometry of the reaction to calculate the amount of gas required based on the molar ratio between the gas and oxygen in the reaction.
When carbon is burned in the presence of oxygen the carbon and oxygen combine to prodon dioxide if you start w 6 grams of carbon and 16 grams of oxygen and they react completely how many grams of car?
the equation of this reaction is C + O2 = CO2 (s) (g) (g) according to this reaction 1mol of Carbon react with same number of moles of Oxygen. and this combination gives 1mol of CO2 as result. so first we have to find the number of moles in the weight we are going to use. 12 grams of carbon contains 1mol so 6 grams of carbon contains ( 1/12 )*6 = 0.5mol 32 grams of Oxygen contains 1mol 16 grams of Oxygen contains ( 1/36 )*16 = 0.5mol as 1mol of C react with 1mol of O2,0.5mol of C react with 0.5mol of O2. this produces equal number of moles (0.5mol) of CO2.
How are helium and oxygen different?
Helium is completely inert. Oxygen will react with many substances.
What is the mole ratio of oxygen to pentane?
The mole ratio of oxygen to pentane in the balanced chemical equation for the combustion of pentane is 13:1. This means that 13 moles of oxygen are required to completely react with 1 mole of pentane.
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.
