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3 years ago

How do you solve stoichiometry problems

1 answer:

nataly862011 [7]3 years ago

7 0

Ok so the way I do it is as simple as possible.
Firstly look at the reactants and products ( there can be one reactant and one product or more ) you will usually be given the moles of the reactant or products, if you are given grams you can convert into moles by this convertion ( grams/R.M.M ) where R.M.M is the relative atomic mass of your substance ( the mass number of all of the elements in your substance).

Ok when you have moles now look at the ratio between the products and reactants. Usually you will won't know the moles of one substance therefore you will be asked to find moles or mass of that substance.

For example:

When 16 grams of oxygen and 1 gram of hydrogen gas react to produce water. Find the number of grams of water being produced.

O2 + 2H2 -> 2H2O
16g 2g xg

Here we're told the mass of the reactants. In stoichiometry we need to work with moles therefore you need to calculate moles of the reactants.
Firstly find the R.M.M of each reactant.
R.M.M of O2 is 16+16=32 since it's diatomic we add atomic masses of two oxygen atoms.
R.M.M of H2 is 1+1=2, it's also diatomic. (Diatomic two atoms of the same element are joined together). (Ignore the number 2 in front of H2, this number shows us the ratio relationship between reactans or products, i.e when we balance an equation.)

Ok so now find moles:

We have 16 grams of O2
16/R.M.M
16/32 = 0.5 moles

We have 2 grams of H2
1/R.M.M
2/2 = 1 mole
Now back to the equation.

O2 + 2H2 -> 2H2O
0.5 moles 1mole xmoles (it's xmoles because we don't know molarity of water that's what we have find firstly in order to find grams.)
Now look at the ratio between any reactant and product i.e you can choose which reactant to compare to the product, it doesn't make a different ( I will do two or you can do two at the same time)

1st method:
Look at the ratio between O2 and H2O from the reaction above we see the ratio is 1:2 therefore for every 0.5 moles of O2 you get 1 mole of H2O.
1:2
0.5 : x
0.5*2 = 1

2nd method;
Look at the ratio betweem H2 and H2O from the reaction above we see the ratio is 2:2 or 1:1. We have 1 mole of H2 there we must have 1 mole of H2O. We see this is true as both methods give us 1 mole of H2O.

3rd method ( combined):
Look at the ratio between O2, H2 and H2O.
We see that the ratio is 1:2:2
So we have 0.5:1:x
If we multiply 0.5 *2 it equals 1 mole
If we multiply 1*1 we get 1 moles.
Any method is correct and it's up to you to find a comfortable way.
We're not finished in the question we are asked for the mass of water.
So just multiply the number of moles (1mole) by R.M.M of H2O.
1 * R.M.M
R.M.M of H2O = 1+1+16=18
1*18= 18 grams.
And you're finished.

I am sorry if this is so long I want you to understand as much as possible.
In stoichiometry you can also be asked about the empirical formula of a substance. I can show you how do it. If you have any question just tell me.
Hope this helps :).

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An aqueous CsCl solution is 8.00 wt% CsCl and has a density of 1.0643 g/mL at 20°C. What is the boiling point of this solution?

umka2103 [35]

<u>Answer:</u> The boiling point of solution is 100.53

<u>Explanation:</u>

We are given:

8.00 wt % of CsCl

This means that 8.00 grams of CsCl is present in 100 grams of solution

Mass of solvent = (100 - 8) g = 92 grams

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

Or,

$\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Boiling point of pure solution = 100°C

i = Vant hoff factor = 2 (For CsCl)

$K_b$ = molal boiling point elevation constant = 0.51°C/m

$m_{solute}$ = Given mass of solute (CsCl) = 8.00 g

$M_{solute}$ = Molar mass of solute (CsCl) = 168.4 g/mol

$W_{solvent}$ = Mass of solvent (water) = 92 g

Putting values in above equation, we get:

$\text{Boiling point of solution}-100=2\times 0.51^oC/m\times \frac{8.00\times 1000}{168.4g/mol\times 92}\\\\\text{Boiling point of solution}=100.53^oC$

Hence, the boiling point of solution is 100.53

6 0

3 years ago

an engineer wishes to design a container that will hold 12.0 mol of ethane at a pressure no greater than 5.00x10*2 kPa and a tem

OleMash [197]

Answer:

The minimum volume of the container is 0.0649 cubic meters, which is the same as 64.9 liters.

Explanation:

Assume that ethane behaves as an ideal gas under these conditions.

By the ideal gas law,

$P\cdot V = n\cdot R\cdot T$ ,

$\displaystyle V = \frac{n\cdot R\cdot T}{P}$ .

where

$P$ is the pressure of the gas,
$V$ is the volume of the gas,
$n$ is the number of moles of particles in this gas,
$R$ is the ideal gas constant, and
$T$ is the absolute temperature of the gas (in degrees Kelvins.)

The numerical value of $R$ will be $8.314$ if $P$ , $V$ , and $T$ are in SI units. Convert these values to SI units:

$P =\rm 5.00\times 10^{2}\;kPa = 5.00\times 10^{2}\times 10^{3}\; Pa = 5.00\times 10^{5}\; Pa$ ;
$V$ shall be in cubic meters, $\rm m^{3}$ ;
$T = \rm 52.0 \textdegree C = (52.0 + 273.15)\; K = 325.15\; K$ .

Apply the ideal gas law:

$\displaystyle \begin{aligned}V &= \frac{n\cdot R\cdot T}{P}\\ &= \frac{12.0\times 8.314\times 325.15}{5.00\times 10^{5}}\\ &= \rm 0.0649\; m^{3} \\ &= \rm (0.0649\times 10^{3})\; L \\ &=\rm 64.9\; L\end{aligned}$ .