Please do not block ads on this website. The mole ratio is the stoichiometric ratio of reactants and products and is the ratio of the stoichiometric coefficients for reactants and products found in the balanced chemical equation. Examples of Mole Ratios (stoichiometric ratios)
Do you know this? Join AUS-e-TUTE! Play the game now! For the balanced chemical reaction shown below: 2Mg(s) + O2(g) → 2MgO(s) the mole ratio of Mg : O2 : MgO is 2:1:2 That is, it requires 2 moles of magnesium and 1 mole of oxygen to produce 2 moles of magnesium oxide. If only 1 mole of magnesium was present, it would require 1 ÷ 2 = ½ mole of oxygen gas to produce 2 ÷ 2 = 1 mole magnesium oxide. If 10 moles of magnesium were present, it would require (1 ÷ 2) × 10 = 5 moles of oxygen gas to produce (2 ÷ 2) × 10 = 10 moles of magnesium oxide. For n moles of magnesium :
The table below shows the moles of MgO produced when various amounts of Mg in moles react with the stoichiometric ratio of O2:
Do you understand this? Join AUS-e-TUTE! Take the test now! It is possible to calculate the mass of each reactant and product using the mole ratio (stoichiometric ratio) from the balanced chemical equation and the mathematical equation moles = mass ÷ molar mass For the balanced chemical equation shown below: 2Mg(s) + O2(g) → 2MgO(s) Given a mass of m grams of magnesium:
Can you apply this? Join AUS-e-TUTE! Take the exam now! The Question: 12.2 g of magnesium metal (Mg(s)) reacts completely with oxygen gas (O2(g)) to produce magnesium oxide (MgO(s)). Calculate the mass of oxygen consumed during the reaction and the mass of magnesium oxide produced. How to Answer the Question: (1) Write the balanced chemical equation for the chemical reaction: 2Mg(s) + O2(g) → 2MgO(s) (2) Determine the mole ratio (stoichiometric ratio) from the equation, Mg : O2 : MgO moles(Mg) : moles(O2) : moles(MgO) is 2:1:2 (3) Use the mole ratios to calculate the mass of O2 consumed and MgO produced as shown below:
Moderators: Chem_Mod, Chem_Admin
Sophia Diaz - Dis 1B Posts: 73 Joined: Fri Apr 06, 2018 11:02 am Been upvoted: 1 time
One of the questions I keep running into is: Can the mass of the products of an equation be more than the mass of the reactants?I can see the answer being yes because maybe in certain reactions surrounding elements during the experiment like oxygen may affect the reaction, but I don't know if that makes that oxygen a reactant. But then again, logically, I can see the answer being no since it makes sense that you can't have more mass than you started with. So in this case, is the answer: no, yes, or under certain circumstances?
ErinKim1I Posts: 31 Joined: Fri Apr 06, 2018 11:03 am
I think this question is just testing if you understand law of conservation of mass, so I think the answer should be no.
Jessica Urzua-1H Posts: 37 Joined: Tue Nov 14, 2017 3:01 am
The Law of Conservation of Mass states that matter cannot be created or destroyed. You can conclude that the answer is no, because the mass must be the same for both the reactants and the products no matter what the reactants and products are, including oxygen. It can be tricky because it is sometimes understood that mass of products is supposed to be more than the mass of reactants because products are supposed to be new compounds made from the reactants. Although the reactants make the products, that does not mean that the mass of the products should be any different than the mass of the reactants. This is why we balance equations to make sure the mass and the ratios from the chemical reaction are accurate. I hope this helps!!
Erin Li 1K Posts: 30 Joined: Fri Apr 06, 2018 11:03 am
No; based on the law of conservation of mass, the amount of products is equal to the mass of the reactants since this mass is not created nor destroyed in chemical reactions. They are just bonded differently.
Luis Avalos 1D Posts: 30 Joined: Wed Nov 22, 2017 3:00 am
Like everyone else is saying, chemistry strictly follows the law of conservation of mass. What we would ideally observe is the same masses, but sometimes mass is lost when we transfer the products elsewhere. Return to “Limiting Reactant Calculations”
Users browsing this forum: No registered users and 0 guests |