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2 answers:
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Answer:
27%
Explanation:
Hello,
The following information is missing, but I found it: "1.92 g of sodium sulfate is produced from the reaction of 4.9 g of sulfuric acid and 7.8 g of sodium hydroxide" so the undergoing chemical reaction is:
Now, to compute the percent yield, we must first establish the limiting reagent to subsequently determine the theoretical yield of sodium sulfate because the real (1.92g) is already given, thus, we consider the following procedure:
- The moles of sodium hydroxide that completely react with 0.05 moles of sulfuric acid are:
As this number is higher than the previously computed 0.05 moles of available sulfuric acid, one states that the sulfuric acid is the limiting reagent. Now, the theoretical grams of sodium sulfate are found via:
Finally, the percent yield turns out into:
%
Best regards.
Answer:
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to solve for the molarity of the KOH solution by knowing this base react in a 1:1 mole ratio with nitric acid, HNO3; thus, we can write the following equation, as their moles are the same at the endpoint:
Which in terms of molarities and volumes is:
Thus, we solve for the molarity of the base (KOH) to obtain:
Regards!