Answer:
44.8 g
Explanation:
Density = mass / Volume
Mass = density x Volume = 8.96x 5 = 44.8 g
Limiting reactant : O₂
Mass of N₂O₄ produced = 95.83 g
<h3>Further explanation</h3>
Given
50g nitrous oxide
50g oxygen
Reaction
2N20 + 302 - 2N204
Required
Limiting reactant
mass of N204 produced
Solution
mol N₂O :
mol O₂ :
2N₂O+3O₂⇒ 2N₂O₄
ICE method
1.136 1.5625
1.0416 1.5625 1.0416
0.0944 0 1.0416
Limiting reactant : Oxygen-O₂
Mass N₂O₄(MW=92 g/mol) :
Answer:
The type of reaction for the following equation is combustion equation.
Explanation:
Combustion reaction is defined as the chemical reaction in which a hydrocarbon reacts with oxygen gas to produce carbon dioxide gas and water molecule.
The reaction given to us:
When 1 mole of ethane reacts with 7/2 moles of oxygen gas it gives 3 moles of water and 2 moles of carbon dioxide gas.
The type of reaction for the following equation is combustion equation.
Formation reaction is the formation of 1 mole of product from the constituents of the reactant molecules. The mass of oxygen that must react is 182 gm.
<h3>What is mass and molar mass?</h3>
Mass of the substance is the weight while the molar mass of the substance is the addition of the atomic mass of the individual mass of the constituent atoms of the compound or the molecule.
The chemical reaction can be shown as:
From the reaction, it can be said that 3 moles of oxygen are required to produce 2 moles of aluminium oxide, so x moles of oxygen will be required to produce 3.80 moles of aluminium oxide.
Solving for x:
If 1 mol of oxygen is 32 gm then 5.7 moles of oxygen will be 182.4 gm.
Therefore, option D. 182 gm is the mass of oxygen required.
Learn more about moles and molar mass here:
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