| When does a Reaction stop? | -When one of the Reactants is completely Used up.
-Other reactants are now in Excess |
| What is the Reactant called when its Used Up? | -This is given the name, Limited Reactant, as its Limits the amount of Product made |
| How can a Balanced equation give you the mass of a Product from a given Mass of a Reactant | -First balance the equation
-Find the Relative Formula mass of the Reactant of the Product and Reactant you need
-Find how many moles there is of the Substances you already know the mass of
-Using your balanced equation, find how man moles will be made of the other substances
-Use that number of Moles to finally find out the mass |
| Whats the mass of Aluminium Oxide when 135g of Aluminium is burned in Air? Show workings | -First balance out the Equation [4Al + 3O² = 2Al²O³]
-Find the Relative Formula mass of the Reactants and Products you want [Al=27. O²=32] [Al²O³ = 102]
-Find the number of moles in the Substances you already know [Aluminium in this case] [135 divided by 27 = 5 moles]
-Using that, use the Balanced equation to how many moles in Aluminium Oxide [If there is 5 Moles in Aluminium, and there is 4 lots of it, the ratio will be 1:1.25. Since Aluminium Oxide has 2 lots, it will have 2.5 Moles involved ]
-Finally get the mass of Aluminium Oxide [2.5 x 102 = 255 grams] |
| Magnesium oxide can be made by Burning Magnesium in the Air. What mass of Magnesium is needed to make 100 grams of Magnesium Oxide? Show workings | -First write the Balanced equation [2Mg + O² = 2MgO]
-Find the relative Formula mass or Atomic for the Reactants and Products you want [Mg=24.3] [MgO = 40.3]
-Find out the number of Moles of Magnesium Oxide in 100 grams
[100 divided by 40.3 = 2.48.]
-Find the Ratio of Moles to find the number of moles, compared to the number of moles Made [2 Moles of MgO are made from 2 Moles of Magnesium. So 2.48 moles of MgO can make 2.48 moles of Mg...
-Finally, use that Ratio to find the mass of Magnesium
[2.48 x 24.3 = 60.3] |
| How can you find the Balanced Eqaution, if you know the masses its reactants and products of a Reaction? | -First, divide the Mass of each Sub stance by its Relative Formula Mass to gain the number of Moles
-Then divide the number of moles of each of the substances, by the Smallest numbers of Moles in the Reaction
-Then, if you need to, multiply all the numbers by the same amount to make them Whole Numbers
-Finally, write your balanced symbol equation by putting the Numbers in front of the Formulas |
| Michael burns a Metal, X in oxygen. There is a single product, an oxide of the Metal. Given that 25.4 grams of X Burns in 3.2 grams of Oxygen, write a Balanced Equation for this Reaction. Show workings
Ar of Metal X: 63.5
Mr of X Oxide: 143.0 | 1. Find mass of Metal Oxide made. This is the Only Product so this can mean the Mass of the Metal Oxide can be equals to the Total Mass of the Reactants [25.4g+3.4g = 28.6g]
2. Divide the Mass of each Substance by its Ar or Mr
[Metal X=25.4 / 63.5 = 0.40m] [O2= 3.2/32=0.10]
[X Oxide = 28.6/143 = 0.20]
3. Divide by the number of the Smallest Moles [Here its 0.10]
[O.4/0.1=4] [0.10/0.10=1] [0.20/0.10=2.0
4. Numbers are back to Whole Numbers, so you can write now the Balanced Equation [4X + O²= 2X²O]
[You get the Oxide because of the Metal X reacting with the Oxygen] |
| Muhammad puts 8.14g of Zinc Oxide in a Crucible with 0.30g of Carbon, and was Heated until they reacted. Knowing that the Balance Equation is 2ZnO + C = CO² + 2Zn, Find the Limiting Reactant here | 1. First divide the Mass of each Substance that was Reacted [The Reactants] by its Ar or Mr to find the number of Moles
[ZnO= 8.14/81.4=0.10] [C=0.3/12 = 0.025]
2. Divide by the Smallest number of Moles [0.025 here]
[0.10/0.025 = 4.0] [0.025/0.025 = 1]
3. Using the Rations the Moles, and Compare them with the Products of the Balanced Equation
[Here. ZnO and C React in a ration of 2:1. These masses show that there is a 4:1 ratio of ZnO to C. This would mean that Carbon is the Limiting Factor] |
