Stoichiometry

[xwrathbringerx]

Hi. Each of these 5 questions below is the 1st question of a different exercise and thus should require different types of working out. Could someone please demonstrate how to do each one so that I am able to complete the exercises? Thanx!

1. In an oxyacetylene torch used for cutting steel and welding, acetylene, C2H2, reacts with oxygen to form carbon dioxide and water.
a) Write a balanced equation for this reaction.
b) In such a torch, how many moles of oxygen must be provided per mole of acetylene used, and how many moles of carbon dioxide are formed>
c) How many moles of oxygen are needed and how many moles of carbon dioxide are formed when 0.018 mol of acetylene reactes?

2. A student spilt 18.6 g hydrochloric acid on the bench. How much bicarbonate of soda needs to be sprinkled on this to neutralise it? The reaction produces carbon dioxide gas, sodium chloride and water.

3. If iron (III) oxide is heated with carbon (coke), the products are carbon monoxide and elemental iron. This very roughly approximates what happens in a blast furnace for commercially extracting iron from its ore. How much iron is obtained if 3.2 g iron (III) oxide is heated with 0.65 g carbon?

4. To protect certain solutions from carbon dioxide, their storage bottles are often fitted with a tube containing granular calcium hydroxide, Ca(OH)2. This reacts with carbon dioxide from the air to form calcium carbonate (CaCO3) and water. What volume of carbon dioxide (at STP) can 1.0 g calcium hydroxide absorb?

5. Hydrogen reacts with oxygen to form water. If (a) 100 mL hydrogen and 40 mL oxygen (b) 100 mL hydrogen and 70 mL oxygen, are mixed at room temperature and then ignited to set off the reaction, what will be the final volume of gas (at the same temperature anmd pressure)? Water forms as a liquid at room temeprature and has a negligible volume compared with that of any gases involved.

P.S. I'm sorry if it's a lot but I'm kinda desperate :o

[Russ]

1. In an oxyacetylene torch used for cutting steel and welding, acetylene, C2H2, reacts with oxygen to form carbon dioxide and water.
a) Write a balanced equation for this reaction.
b) In such a torch, how many moles of oxygen must be provided per mole of acetylene used, and how many moles of carbon dioxide are formed>
c) How many moles of oxygen are needed and how many moles of carbon dioxide are formed when 0.018 mol of acetylene reactes?


In stoichiometry, the golden rule is: what you WANT TO KNOW divided by what you KNOW. Using that, you can work out the correct ratio. When you look at an equation, the numbers of the reactants tell you how many moles of those reactants will combine to make the moles of product shown.

a) We're told the products and reactants, so all we need to do is balance. Double the amount of CO2 to balance carbon, no change needed for hydrogen, then you need 2.5O2 molecules to balance oxygen. Since we don't want fractions in equations, double everything to give you:

2C2H2 + 5O2 -> 4CO2 + 2H2O

b)
2 moles of acetylene requires 5 moles of oxygen to react fully. 1 mole of acetylene therefore requires 5/2 moles of oxygen.
2 moles of acetylene produces 4 moles of CO2. 1 mole of acetylene produces 2 moles of CO2

c) .018 mole of C2H2 requires .018 x (5/2) moles of oxygen to react with. (.045 mol of oxygen is the answer)
.018 mole of C2H2 will produce .018 x (4/2) moles of CO2. (.036 mol of carbon dioxide is the answer)


2. A student spilt 18.6 g hydrochloric acid on the bench. How much bicarbonate of soda needs to be sprinkled on this to neutralise it? The reaction produces carbon dioxide gas, sodium chloride and water.


First we need to write a reaction. As it turns out, this doesn't require balancing, making stoichiometry very easy :)

HCl + NaHCO3 -> CO2 + NaCl + H2O

Now, we know that we have 18.6g of HCl. We need to convert this to mole, and n = m/M = 18.6/36.5 = .51 mol of HCl
To "neutralise it" we will need to add the same amount of sodium bicarbonate (mole ratios are the same).

.51 mol of NaHCO3 can be converted to mass (g) by rearranging the above formula of n = m/M
m = n x M = .51 x 84 = 42.84g

Will do the rest when i get back from dinner and have some time.

[Sonic]

3. If iron (III) oxide is heated with carbon (coke), the products are carbon monoxide and elemental iron. This very roughly approximates what happens in a blast furnace for commercially extracting iron from its ore. How much iron is obtained if 3.2 g iron (III) oxide is heated with 0.65 g carbon?

Bearing in mind the golden rule as written by Russ in the last post
Also when converting between moles and grams always use the values given to you in your periodic table - they may be slightly different from what people use here.

Write out an equation - the (III) means the iron has an oxidation state of +3 which, for the purposes of this question, can be likened to having a +3 charge on it

Fe2O3 + C ----> Fe + CO.

This equation will need balancing and it balances out to
Fe2O3 + 3C --> 2Fe + 3CO

Moles Fe2O3 = 3.2/(2*56+3*16) = 0.02moles
Moles C = 0.65/12 = 13/240moles

Since there are 4 moles on the left and 5 on the right it means that for every 4 moles of reactant, 5 moles of products are produced.

Moles Products = (0.02+13/240) x 5/4 = 89/960 moles.

These will be distributed in the ratio of 2 to 3 of Fe and CO respectively:

moles Fe = 89/960 x 2/5 = 89/2400
mass Fe = moles Fe x Mr Fe = 89/2400 x 56 = 2.08g

You should always check to make sure your final answer is reasonable. In this case it is reasonable for we can't get more than we put in and some would be lost due to driving off the oxygen.

4. To protect certain solutions from carbon dioxide, their storage bottles are often fitted with a tube containing granular calcium hydroxide, Ca(OH)2. This reacts with carbon dioxide from the air to form calcium carbonate (CaCO3) and water. What volume of carbon dioxide (at STP) can 1.0 g calcium hydroxide absorb?

Ca(OH)2 + CO2 ----> CaCO3 + H2O
This equation is balanced already :)

We can also work out the amount of moles of Ca(OH)2 by dividing 1.0g by it's molecular mass (100g/mol)

n = 1.0/(40+12+3*16) = 0.01mol

From the equation we can see that one mole of Ca(OH)2 can absorb one mole of CO2 therefore 0.01mol of CO2 can be absorbed.

Let us recall the fact that 1 mol of any ideal gas (and we can assume CO2 is in this case) occupies a volume of 22.4dm^3 at STP.

As one mole occupies 22.4L then it follows that 0.01 mole occupies 22.4/100 = 0.224L = 224cm^3. Mathematically:

0.01/1 = V/22.4 , V = 0.224L = 224cm^3


5. Hydrogen reacts with oxygen to form water. If (a) 100 mL hydrogen and 40 mL oxygen (b) 100 mL hydrogen and 70 mL oxygen, are mixed at room temperature and then ignited to set off the reaction, what will be the final volume of gas (at the same temperature anmd pressure)? Water forms as a liquid at room temeprature and has a negligible volume compared with that of any gases involved.

P.S. I'm sorry if it's a lot but I'm kinda desperate :o

Write the balanced equation 2H2 + O2 ---> 2H2O

If we assume they are ideal gases and so it would follow that
V2/V1 = n2/n1. where n1,V1 are properties of hydrogen and n2,V2 are properties of Oxygen.

From the equation we can see that n1/n2 = 2 so V1/V2 = 2.

a) Since V2 = 40ml it means that 80ml of V1 is used.

However, we have 100mL of H2 so 20mL will not react (it is the excess) and because the reaction goes to completion because water has a negligible volume the answer is 20mL.

b) V1 is now the limiting reagent. If we put this into the equation we get

100/V2 = 2 and so V2=50mL

For the same reasons as above we can say that the answer is the total of O2 minus the O2 that reacts:
70-50 = 20mL

[xwrathbringerx]

Thank you so much!!!!:D