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A portland cement sample contained 20% SiO2 by weight derived from two silicate compounds, SiO2.2CaO and SiO2.3CaO that are present in the cement in the mole ratio 3 : 4. Determine the percent by weight of each silicate compound in the cement.

Answer key from book : 24.57%, 43.43%

My answer: 36.03%, 63.09%

Assuming the first silicate compound as W1 & the second one as W2 and their overall composition contribution to the sample as 100%. I got one equation W1+W2= 357.4 and using the mole ratios given, I equated them to get W1= .564 *W2. upon resubstituting that into the first equation, I got 128.8 g for W1 and 228.51 g for W2. Calculating the percentages for each, I got 36.03% and 63.9 % respectively. –

Note: I do find the percentages from the answer key weird too as they don't quite indicate a 100% composition contribution from the constituents. Or, maybe I am just missing something.

The help is much appreciated.

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    $\begingroup$ You should include you solution as well.l, not just your result. $\endgroup$
    – Poutnik
    Commented Feb 9, 2020 at 17:40
  • $\begingroup$ Ok. Assuming the first silicate compound as W1 & the second one as W2 and their overall composition contribution to the sample as 100%. I got one equation W1+W2= 357.4 and using the mole ratios given, I equated them to get W1= .564 *W2. upon resubstituting that into the first equation, I got 128.8 g for W1 and 228.51 g for W2. Calculating the percentages for each, I got 36.03% and 63.9 % respectively. $\endgroup$
    – abhi das
    Commented Feb 9, 2020 at 17:46
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    $\begingroup$ Technical note: The question improvement should be done in the question itself, not in the comments. Factual note: There is not said W1&W2 contribution is 100%. You also do not respect their ratio 3:4. $\endgroup$
    – Poutnik
    Commented Feb 9, 2020 at 17:51
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    $\begingroup$ Yes, I think this is the main mistake. w1+w2 is not equal to the weight of cement. $\endgroup$
    – AChem
    Commented Feb 9, 2020 at 17:52
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    $\begingroup$ SE is not just a forum, questions and answers should be written in a way so they can be found and used later by someone else, who gets stuck with a similar problem. Instead of trying to be ironic, you should rather do something about it. $\endgroup$
    – Karl
    Commented Feb 9, 2020 at 22:28

1 Answer 1

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This problem seems to have wrong data ! Look ! The molar mass of $Ca_2SiO_4$ is $172$. The molar mass of $Ca_3SiO_5$ is $228$. If you mix $3$ moles $Ca_2SiO_4$, (= $516$ g), and $4$ moles $Ca_3SiO_5$, (= $912$ g), the total weight is : $516 + 912 = 1428$ g. This mixture contains $3+4 = 7$ moles of $SiO_2$, which weighs $420$ g. In this mixture, the ratio $SiO_2$/total mass = $420/1428$ = $29.4$%. Well. This is contrary to the initial data saying that the proportion of $SiO_2$ is $20$% !!! Where is the mistake or the misconception ?

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  • $\begingroup$ Thanks for your reply Maurice. I have a link here that will direct you to site I retrieved the question from. I just rechecked it and it has been presented as is. Pg 72, problem 3.44 - vdocuments.site/stoichiometry-and-process-calculations-1.html $\endgroup$
    – abhi das
    Commented Feb 9, 2020 at 22:23
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    $\begingroup$ No, I am afraid the above calculation is not correct. Why are you assuming the total weight of calcium silicates is the total weight of cement? $\endgroup$
    – AChem
    Commented Feb 10, 2020 at 1:59

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