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Balanced Symbol Equation

4,420 bytes added, 14:01, 3 January 2019
Key Stage 4
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===Limiting Reactants===
: A [[Limiting Reactant|limiting reactant]] is a [[reactant]] that is not supplied in a large enough quantity for a complete [[Chemical Reaction|reaction]] with the other [[reactant]]s.
 
2H<sub>2</sub>(g) + O<sub>2</sub>(g) → 2H<sub>2</sub>O(l)
 
: If there is 4g of [[Hydrogen]] then 32g of [[Oxygen]] is needed for a complete [[Chemical Reaction|reaction]] to occur.
: If there is 4g of [[Hydrogen]] but only 31g of [[Oxygen]] then the [[Oxygen]] is a [[Limiting Reactant|limiting reactant]].
: If there is 3g of [[Hydrogen]] but only 32g of [[Oxygen]] then the [[Hydrogen]] is a [[Limiting Reactant|limiting reactant]].
 
{| class="wikitable"
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| style="height:20px; width:200px; text-align:center;" |If there are are 6g of [[Carbon]] and 15g of [[Oxygen]] in the following [[Chemical Reaction|reaction]]:
C + O<sub>2</sub> → CO<sub>2</sub>
Which is the [[Limiting Reactant|limiting reactant]] in this [[Chemical Reaction|reaction]]?
 
| style="height:20px; width:200px; text-align:center;" |If there are are 27g of [[Aluminium]] and 32g of [[Oxygen]] in the following [[Chemical Reaction|reaction]]:
4Al(s) + 3O<sub>2</sub>(g) → 2Al<sub>2</sub>O<sub>3</sub>(s)
Which is the [[Limiting Reactant|limiting reactant]] in this [[Chemical Reaction|reaction]]?
 
| style="height:20px; width:200px; text-align:center;" |If there are are 180g of [[Glucose]] and 200g of [[Oxygen]] in the following [[Chemical Reaction|reaction]]:
C<sub>6</sub>H<sub>12</sub>O<sub>6</sub>(s) + 6O<sub>2</sub>(g) → 6H<sub>2</sub>O(l) + 6CO<sub>2</sub>(g)
Which is the [[Limiting Reactant|limiting reactant]] in this [[Chemical Reaction|reaction]]?
 
|-
| style="height:20px; width:200px; text-align:center;" |
Find the [[Relative Formula Mass]] of the [[reactant]]s.
 
M<sub>r</sub> of C = 12g
 
M<sub>r</sub> of O<sub>2</sub> = 32g
 
| style="height:20px; width:200px; text-align:center;" |
Find the [[Relative Formula Mass]] of the [[reactant]]s.
 
M<sub>r</sub> of NaOH = 40g
 
M<sub>r</sub> of HCl = 36.5g
 
| style="height:20px; width:200px; text-align:center;" |
Find the [[Relative Formula Mass]] of the [[reactant]]s.
 
M<sub>r</sub> of NaOH = 40g
 
M<sub>r</sub> of HCl = 36.5g
|-
| style="height:20px; width:200px; text-align:center;" |
State the [[ratio]] of [[mole]]s of each [[chemical]] needed.
 
1 [[mole]]s of C is needed for every 1 [[mole]] of O<sub>2</sub>
 
| style="height:20px; width:200px; text-align:center;" |
State the [[ratio]] of [[mole]]s of each [[chemical]] needed.
 
4 [[mole]]s of Al are needed for every 3 [[mole]]s of O<sub>2</sub>
 
| style="height:20px; width:200px; text-align:center;" |
State the [[ratio]] of [[mole]]s of each [[chemical]] needed.
 
1 [[mole]] of C<sub>6</sub>H<sub>12</sub>O<sub>6</sub> is needed for every 6 [[mole]]s of O<sub>2</sub>
|-
 
| style="height:20px; width:200px; text-align:center;" |
Find the number of [[mole]]s supplied of each [[chemical]].
 
No. [[Mole]]s of C = <math>\frac{Mass}{M_r}</math>
 
No. [[Mole]]s of C = <math>\frac{6}{12}</math>
 
No. [[Mole]]s = 0.5 mol
 
Therefore 0.5 [[mole]]s of O<sub>2</sub> is needed.
 
No. [[Mole]]s of O = <math>\frac{Mass}{M_r}</math>
 
No. [[Mole]]s of O = <math>\frac{15}{32}</math>
 
No. [[Mole]]s = 0.46 mol
 
[[Oxygen]] is the [[Limiting Reactant|limiting reactant]].
1 [[mole]] of O<sub>2</sub> = 32g
| style="height:20px; width:200px; text-align:center;" |
Find the number of [[mole]]s supplied of each [[chemical]].
 
No. [[Mole]]s of Al = <math>\frac{Mass}{M_r}</math>
 
No. [[Mole]]s of Al = <math>\frac{27}{27}</math>
 
No. [[Mole]]s = 1 mol
 
Therefore 0.75 [[mole]]s of O<sub>2</sub> are needed.
 
No. [[Mole]]s of O = <math>\frac{Mass}{M_r}</math>
 
No. [[Mole]]s of O = <math>\frac{32}{32}</math>
 
No. [[Mole]]s = 1 mol
 
[[Oxygen]] is the [[Limiting Reactant|limiting reactant]].
 
| style="height:20px; width:200px; text-align:center;" |
Find the number of [[mole]]s supplied of each [[chemical]].
 
No. [[Mole]]s of C<sub>6</sub>H<sub>12</sub>O<sub>6</sub> = <math>\frac{Mass}{M_r}</math>
 
No. [[Mole]]s of C<sub>6</sub>H<sub>12</sub>O<sub>6</sub> = <math>\frac{180}{180}</math>
 
No. [[Mole]]s = 1 mol
 
Therefore 6 [[mole]]s of O<sub>2</sub> are needed.
 
No. [[Mole]]s of O = <math>\frac{Mass}{M_r}</math>
 
No. [[Mole]]s of O = <math>\frac{200}{32}</math>
 
No. [[Mole]]s = 6.25 mol
 
[[Glucose]] is the [[Limiting Reactant|limiting reactant]].
 
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4Al(s) + 3O<sub>2</sub>(g) → 2Al<sub>2</sub>O<sub>3</sub>(s)
===Balancing Equations===