Chem q (1 Viewer)

[=)(=]

 

[username_2]

View attachment 35006

Well, just hints but:

1) Write the chemical equation. Find the number of moles of each reactant in the reaction using n = m/MM
2) If one is in excess, that means not all of the moles of this excess chemical is used up. (i.e. if AlCl3 is in excess, then only the number of moles of NaOH reacts to form the precipitate).
3) Using the reaction equation, use the molar ratios to equate how many moles of the product is formed. Use m = n*MM to find the mass of the precipitate.

Hope it helps.

 

[Octavius]

Is the answer around 7.8g?

 

[=)(=]

Is the answer around 7.8g?

yes

 

[=)(=]

Well, just hints but:

1) Write the chemical equation. Find the number of moles of each reactant in the reaction using n = m/MM
2) If one is in excess, that means not all of the moles of this excess chemical is used up. (i.e. if AlCl3 is in excess, then only the number of moles of NaOH reacts to form the precipitate).
3) Using the reaction equation, use the molar ratios to equate how many moles of the product is formed. Use m = n*MM to find the mass of the precipitate.

Hope it helps.

I tried that but for some reason i got something like 0.01

 

[Octavius]

I tried that but for some reason i got something like 0.01

Ok I will try and give a sort of method:
1. Calculate the number of moles for the reactants
2. Find the limiting reagent
To do this I will use this example:
The mole ration between the reactants in 1:3
For n moles of AlCl3 the moles of NaOH required is:
number of actual moles of AlCl3/stoichiometric moles of AlCl3( which is 1) * stoichiometric moles of NaOH (which is 3)
3. Once you have found the limiting reagent (NaOH) work out the mole ratio between it and the solid. This is 3:1. Therefore, divide the actual moles of NaOH by 3 to get the actual moles of Al(OH)3.
4. Calculate the mass using n=m/mm

 

[Octavius]

Hope it helps

 

[=)(=]

I got it now thxs guys, also when doing this should i leave the mole caluclation in exact form and also is there a faster way then converting to moles?

 

[=)(=]

also have this q that i cant get

 

[jimmysmith560]

also have this q that i cant get View attachment 35007

Would the following approach help? Note that percentages have been modified slightly:

1. Find the empirical formula of a compound that has 40.87% of Carbon (C), 21.98% of Chlorine (Cl), 3.72% of Hydrogen (H), 8.67% of Nitrogen (N), 24.77% of Oxygen (O).
2. Find molar mass of each component: C=12.0107, Cl=35.453, H=1.00794, N=14.0067, O=15.9994
3. Convert to moles: C=3.4027991707394, Cl=0.61997574253237, H=3.6906958747544, N=0.61898948360428, O=1.5481830568646
4. Find smallest mole value: 0.61898948360428
5. Divide all components by the smallest value: C=5.4973456914413, Cl=1.0015933371313, H=5.9624532766924, N=1, O=2.5011459772302
6. Divide by the fractional components (0.49734569144128, 0.98854917072583) of each mole value: C=11.181405858895, Cl=2.0372052689838, H=12.127418165677, N=2.0339644778575, O=5.0872420716224
7. Round to closest whole numbers: C=11, Cl=2, H=12, N=2, O=5
8. Combine to get the empirical formula: C11Cl2H12N2O5

To find its relative molecular mass, you should follow these steps:

1. Determine the molecular formula of the molecule (which we now have).
2. Use the periodic table to determine the atomic mass of each element in the molecule.
3. Multiply each element's atomic mass by the number of atoms of that element in the molecule. This number is represented by the subscript next to the element symbol in the molecular formula.
4. Add these values together for each different atom in the molecule.

#​	Atom​	Molar Mass​	Subtotal Mass​
11​	C​	12.01​	132.12​
2​	Cl​	35.45​	70.91​
12​	H​	1.01​	12.09​
2​	N​	14.01​	28.01​
5​	O​	16.00​	80.00​

The total will be the molecular mass of the compound, i.e. 323.13.

 

[=)(=]

I am not sure about the second step

 

[jimmysmith560]

I am not sure about the second step

You can also search for the molar mass of each element online.

Getting the number of moles in terms of each element involves treating the percentages as grams (out of 100 g) then dividing each of them by their respective molar mass, for instance:

as seen above for carbon.

 

[=)(=]

You can also search for the molar mass of each element online.

Getting the number of moles in terms of each element involves treating the percentages as grams (out of 100 g) then dividing each of them by their respective molar mass, for instance:

as seen above for carbon.

Ahh i see now thank you