To solve this problem you must first write a balanced reaction. Then use stoichiometry and the Ideal Gas Law to solve for the answer.
First, write the balanced reaction. The reactants are Mg and O2, and the product is MgO. For step by step instructions on how to write a balanced reaction, see the Related Questions to the left.
Then find the number of mole of oxygen in the container using the Ideal Gas Law. For step-by-step instructions on solving Ideal Gas Law problems, see the Related Questions to the left.
Finally, use stoichiometry to determine how much Mg will react with that amount of O2. And yes, you guessed it, see the Related Questions to the left for how to do that.
.525mol
Propane is C3H8 and the combustion equation is C3H8 + 5O2 ==> 3CO2 + 4H2OSo the complete combustion of 1 mole of propane requires 5 moles of oxygen.
The balanced equation for the reaction is 2 C4H10 + 13 O2 -> 8 CO2 + 10 H2O. This shows that 13 moles of diatomic oxygen are required to burn 2 moles of butane. By proportionality, (4.8/2)13 or 31.2 moles of oxygen are required to burn 4.8 moles of butane. This corresponds to 31.2(32) or 1.0 X 103 grams of oxygen.
the formula is no. moles is mass / molecular mass. As the number of moles is 1, the mass required will be exactly the same as the molecular mass, which is 58.32g
To answer this question, it is necessary to have an equation for the reaction. The most common such reaction is complete combustion, which follows the equation: C2H4 + 3 O2 = 2 CO2 + 2 H2O. This equation shows that 3 moles of diatomic oxygen are required to react completely with one mole of C2H4. Therefore, for 1.50 moles of C2H4, 3 X 1.5 = 4.50 moles of oxygen will be required. Oxygen is close to an ideal gas at standard temperature and pressure. Each mole of ideal gas at stp occupies 22.4 liters. Therefore, 4.50 X 22.4 = 101 liters of oxygen, to the justified number of significant digits, will be needed.
10 moles of oxygen atoms or 5 moles of oxygen molecules.
No moles of oxygen are produced by complete combustion of propane. Oxygen is CONSUMED, not produced. For combustion of 4 moles of propane, it will use 20 moles of oxygen.
The formula for magnesium oxide is MgO, showing that each formula unit of magnesium oxide contains one mole of magnesium ions. Therefore, if there is ample oxygen available, 4 moles of magnesium will form 4 moles of magnesium oxide.
Propane is C3H8 and the combustion equation is C3H8 + 5O2 ==> 3CO2 + 4H2OSo the complete combustion of 1 mole of propane requires 5 moles of oxygen.
From the formula, you have 2 atoms of Magnesium combine with one oxygen molecule to form 2 molecules of magnesium oxide. So when 4 magnesium atoms combine with two molecules of oxygen you get 4 magnesium oxide molecules. So from 4 moles of magnesium you get 4 moles of Magnesium oxide.
C3H8 + 5O2 --> 3CO2 + 4H2O 2.75 mole C3H8 (5 moles O2/1 mole C3H8)(32 grams/1 moleO2) = 440 grams oxygen required =====================
2Mg + O2 -----> 2MgO So two moles of magnesium oxide are formed if x moles of magnesium are allowed to react with only 1 mole of oxygen molecules. The oxygen has become the limiting ingredient.
The ratio of propane to oxygen is 1:5. So for every mole of propane, 5 moles of oxygen gas are required for the complete combustion of propane.Balanced equation:C3H8 + 5O2 --> 3CO2 + 4H2O
Since there are no subscripts, there is .800 moles of Oxygen atoms and .800 moles of Magnesium because there is .800 moles of the compound.
1 mole
Since oxygen is diatomic it requires 2 moles of oxygen.
Balanced Formula:2Mg + O2 --> 2MgOMole ratio:2 : 1 : 2Givens:.486 g oxygen.738 g magnesium24.3 g = atomic mass of magnesium16.0 g = atomic mass of oxygen40.3 g = molecular mass of magnesium oxideFind the amount (in moles) of Magnesium oxide that oneelement will make:(.486 g O) / (16.0 g O) × (2 moles MgO)= .0608 moles MgO(.783 g Mg) / (24.3 g Mg) = .0322 moles MgOThere is less MgO produced with magnesium than oxygen; therefore, magnesium is the limiting reactant and the oxygen is the excess reactant. The magnesium determines how much Magnesium oxide is produced. It would be good to get .0608 moles of MgO, but there isn't enough magnesium. So the amount of MgO produced will be determined on the amount of Magnesium.Convert moles of MgO produced with the amount of oxygen to grams:.0322 mol MgO (40.3 g) = 1.30 grams of MgO produced--------------------------------------------------------------------------------------------------------You will need 3 moles of oxygen if you start with six moles of magnesium. This will allow you to produce 6 moles of magnesium oxide.Source: (e2020)
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