When heated, sodium metal combines near chlorine gas to form solid sodium chloride...?
When heated, sodium metal combines with chlorine gas to form solid sodium chloride:
2NA + Cl2 --> 2NaCl
If 8.30 g of sodium is heated with an excess of chlorine, a total of 19.5 g of sodium chloride is produced. Determine the abstract yield and percentage yield.
I missed the lesson at institution, so im really confused. Can anyone help me please? Thanks.
Answers:
So first you must convert the amount of grams of sodium which you are given into moles of sodium by dividing by the molecular weight of sodium (22.99) since ALL stoichiometry is done on a molar proof (as that's the way balanced equations are written). So, i'll do that below
8.30 g of Na/22.99 g/mol = 0.361 moles
next you look at the balance equations, you see that for each 2 moles of sodium which you consume, you produce 2 moles of sodium chloride (a 1:1 molar ratio). So in a minute, we convert the moles of Sodium into moles of sodium chloride using this relationship
0.361 moles of Na * (2 mols NaCl/2 mol Na) = 0.361 moles of NaCl
That's the theoretical molar yield of NaCl but we want to know how masses GRAMS we're supposed to get, so we do that by multiplying the number of moles by the molecular weight as shown below:
0.361 moles of NaCl * (58.44 g/mol) = 21.1 g of NaCl
So that's what we're supposed to seize, to calculate percent yield, we use the following formula:
% verbs = (actual yield/theoretical yield) * 100
since we actually made 19.5 grams of sodium chloride, our percent yield is calculated as follows:
% abandon = 19.5/21.1 *100 = 92.4%
A simple stoich problem. You just need to remember to other change grams given to moles and work from there.
step 1
Change grams of limiting chemical to moles.
8.30 g Na / 23 g Na per mole Na from the intermittent table = 0.357 mol Na used.
step 2
look at the mole ratio of reactants and products in your balanced equation.
2 Na reactants give you 2 NaCl products, or 2 mol Na to 2 mol NaCl. That means that 0.357 mol Na should give you 0.357 mol NaCl.
step 3
vary moles back to grams of products.
0.357 mol NaCl x 58.5 g NaCl per mol = 20.9 g NaCl
**The theoretical give up is 20.9 g NaCl**
Percent yield = (actual yield / speculative yield) x 100
actual yield is given as 19.5 g NaCl
theoretical let go is calculated to be 20.9 g NaCl
19.5 / 20.9 = 0.933 x 100 = 93.3% yield
**Percent yield is 93.3%** Source(s): Chem Teacher
Related Questions:
What color does silver metal within river form?
Do pure metals form crystals?
Why are metal sulphide ores roasted contained by nouns to form the oxide up to that time mortal reduced to the metal beside carbon?
When iron (III) oxide is heated next to coke, which is pure carbon, iron metal and carbon monoxide are formed.?
What gas is formed when a metal react near an sour such as HNO3?
2NA + Cl2 --> 2NaCl
If 8.30 g of sodium is heated with an excess of chlorine, a total of 19.5 g of sodium chloride is produced. Determine the abstract yield and percentage yield.
I missed the lesson at institution, so im really confused. Can anyone help me please? Thanks.
Answers:
So first you must convert the amount of grams of sodium which you are given into moles of sodium by dividing by the molecular weight of sodium (22.99) since ALL stoichiometry is done on a molar proof (as that's the way balanced equations are written). So, i'll do that below
8.30 g of Na/22.99 g/mol = 0.361 moles
next you look at the balance equations, you see that for each 2 moles of sodium which you consume, you produce 2 moles of sodium chloride (a 1:1 molar ratio). So in a minute, we convert the moles of Sodium into moles of sodium chloride using this relationship
0.361 moles of Na * (2 mols NaCl/2 mol Na) = 0.361 moles of NaCl
That's the theoretical molar yield of NaCl but we want to know how masses GRAMS we're supposed to get, so we do that by multiplying the number of moles by the molecular weight as shown below:
0.361 moles of NaCl * (58.44 g/mol) = 21.1 g of NaCl
So that's what we're supposed to seize, to calculate percent yield, we use the following formula:
% verbs = (actual yield/theoretical yield) * 100
since we actually made 19.5 grams of sodium chloride, our percent yield is calculated as follows:
% abandon = 19.5/21.1 *100 = 92.4%
A simple stoich problem. You just need to remember to other change grams given to moles and work from there.
step 1
Change grams of limiting chemical to moles.
8.30 g Na / 23 g Na per mole Na from the intermittent table = 0.357 mol Na used.
step 2
look at the mole ratio of reactants and products in your balanced equation.
2 Na reactants give you 2 NaCl products, or 2 mol Na to 2 mol NaCl. That means that 0.357 mol Na should give you 0.357 mol NaCl.
step 3
vary moles back to grams of products.
0.357 mol NaCl x 58.5 g NaCl per mol = 20.9 g NaCl
**The theoretical give up is 20.9 g NaCl**
Percent yield = (actual yield / speculative yield) x 100
actual yield is given as 19.5 g NaCl
theoretical let go is calculated to be 20.9 g NaCl
19.5 / 20.9 = 0.933 x 100 = 93.3% yield
**Percent yield is 93.3%** Source(s): Chem Teacher
Related Questions: