M18 hELP!

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M18 hELP!

Postby JennyCKim1J » Fri Oct 06, 2017 8:51 am

The acid H2A (where A stands for an unknown group of atoms) has molar mass 168 g mol 1. H2A reacts with the base XOH (molar mass 125 g mol -1) to produce H2O and the salt X2A. In one experiment, 1.20 g of H2A reacts with 1.00 g of XOH to form 0.985 g of X2A. What is the percentage yield of the reaction?

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Re: M18 hELP!

Postby Sarah_Stay_1D » Fri Oct 06, 2017 10:01 am

Don't let the unknown compound freak you out! First just write out a balanced reaction - H2A + 2XOH -> 2H2O + X2A. Since you already know the molar mass of both H2A and XOH, you can convert the grams to moles. You should get 0.00714 moles of H2A and .008 moles of 2XOH. Since H2A and 2XOH are in a 1:2, we can multiply 0.00714 by 2, and see that 2XOH is the limiting reactant. Now, since there are .008 moles of 2XOH, there should be .004 moles of X2A, since it is a 2:1. To find the molar mass of X2A, you can use the molar masses of H2A and XOH. Just subtract the masses of the compounds we know. So after subtracting hydrogen, there should be 165.98 g/mol of A. After subtracting oxygen and hydrogen there should be 107.992 g/mol of X. now simply calculate the molar mass of X2A (107.992*2 + 165.98 = 273.97). Now convert .004 moles to grams by multiplying (273.97*.004 = 1.096). 1.096 g is the theoretical yield. SO to find the percent yield, divide .985 g by 1.096 g to get 89.9%.


Srbui Azarapetian 2C
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Re: M18 hELP!

Postby Srbui Azarapetian 2C » Fri Oct 06, 2017 10:09 am

The first step to write a balanced equation.
This would be H2A + 2XOH --> 2H20 +X2A
Then with the given masses and molar masses, divide the mass by the molar mass for each of the reactants.
For H2A :
1.20g/168gmol-1 = 0.00714 mol of H2A
For XOH:
1.00g/125gmol-1 = 0.00800 mol of XOH
Then you multiply of their respective stoichiometric ratios, from the balanced chemical equation, respectively to find the moles of X2A produced, depending on the limiting reactant.
0.00714 mol of H2A x 1mol X2A/1mol H2A = 0.00714 moles of X2A
0.00800 mol of XOH x 1mol X2A/ 2mol XOH = 0.00400 moles of X2A

Since XOH yields the least amount of X2A as the limiting reactant, we use 0.00400 moles of X2A.

Next you find the molar mass of X2A to covert to grams, but as both are unidentified atoms, you must subtract from their original given molar masses in their original compounds.
The total molar mass of H2A is 168gmol-1 and we know that the molar mass of Hydrogen is 1.01gmol-1. Therefore,
168-(2x1.01)= 165.98gmol-1 of the A group of atoms

The total molar mass of XOH is 125gmol-1 and we know the molar mass of Hydrogen is 1.01 and Oxygen is 16.00gmol-1. Therefore,
125-16-1.01= 107.99g/mol of the X group of atoms.
Since there are 2 X atoms in X2A and one A, the molar mass of X2A is 2(107.99) + 165.98 = 273.97 gmol-1
Multiply the molar mass of X2A by the previously found moles.
(273.97 gmol-1) x 0.00400 moles = 1.09588 grams
Percentage yield is actual/theoretical yield x100
In this case, 0.986 g/ 1.09588 g x 100 = 89.9 % percentage yield.

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Re: M18 hELP!

Postby RaviAmin1H » Fri Oct 06, 2017 10:27 am

I think both of the replies above have the wrong mass of X2A

x = 108

so 108* 2 = 216
216+ 166 = 382

unless I am wrong I think that is the proper molar mass that must be pultiplied by .004 mol of X2A

resulting in 1.53 g of X2A

.985/1.53 * 100 = 64.4%

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