I have a few questions on how to do a few problems that I got wrong for homework.
1. If 5.28 g of tin reacted with fluorine to form 8.65 g of a metal fluoride, what is the simplest formula of the fluoride?
I know my biggest problem here was that I misread the problem and didn't see that it was a metal fluoride, so my simplest formula came out to be simply SnF2 [the correct answer is SnF4]. How do I account for the fact that the mass given of the metal fluoride is not just of the fluoride?
2. What is the total mass of products formed when 19.0 g of carbon disulfide is burned in air? What mass of carbon disulfide would have to be burned to produce a mixture of carbon disulfide and sulfur dioxide that masses 34.4 g?
CS2 + 3O2 -> CO2 + 2SO2
I got the first part right, there 43.0 g total of products.
I wasn't sure how to do the second question though.
Also, a question asked to write the formula of nitrogen oxide. After looking this up, I found that this is a general term given to any oxide of nitrogen. Why would this be wrong? N2O3
Last one:
Aspirin, C9H8O4, is prepared by reacting salicylic acid, C7H6O3, with acetic anhydride, C4H6O3, in the reaction:
C7H6O3(s) + C4H6O3(l) -> C9H8O4(s) + C2H4O2 (l)
A student is told to prepare 45.0 g of aspirin. She is told to use a 55.0% excess of acetic anhydride and to expect to get an 85.0% yield in the reaction. How many grams of each reaction should she use?
What's really tripping me up here is the percent excess in yield; i'm not sure how to incorporate it into the problem. Could someone give me a rough outline of what needs to be done?
One question per post please.
The mass of the metal fluoride is given, ans well as the mass of the metal in it. The mass of the fluoride ion is the difference between them.
Use Ratio and proportion to solve the second part of number 2.
Nitrogen oxide can have many formulas. NO would be the plainest. Normally, chemists call that Nitrogen(II) oxide, or older ancient chemists call it Nitrogen monoxide.
On the percent yield....If she wanted 45g , then use ingredents to get 45/.85 grams.
Post your work on any of this if you get stuck. We can help you through.
52.94 g of C7H6o3
60.65 g of C4H6O3
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Sure, I'd be happy to help you with your questions one at a time. Let's start with the first one:
1. To determine the simplest formula of the metal fluoride, we need to calculate the ratio of the masses of tin and fluorine.
Given:
Mass of tin = 5.28 g
Mass of metal fluoride = 8.65 g
First, we need to find the mass of the fluorine in the metal fluoride. We can do this by subtracting the mass of tin from the mass of the metal fluoride:
Mass of fluorine = Mass of metal fluoride - Mass of tin
Mass of fluorine = 8.65 g - 5.28 g
Mass of fluorine = 3.37 g
Now, we can calculate the ratio of the mass of tin to the mass of fluorine:
Mass ratio = Mass of tin / Mass of fluorine
Mass ratio = 5.28 g / 3.37 g
Simplifying this ratio gives us:
Mass ratio = 1.566
Based on the mass ratio, we can determine the simplest ratio of tin to fluorine atoms in the compound. In this case, the ratio is close to 1.5 (or 3:2).
The simplest formula of the fluoride would be SnF4, since there are four fluorine atoms for every tin atom.
I hope this explanation helps! Let me know if you have any further questions.