Ammonia Synthesis Lab Report

If one wanted to find the volume of mole this gas at STP, then all that is required is to divide the 0.0377 liters at STP by the original number of moles of magnesium from the start of the experiment; this would yield the results at STP if one mole of magnesium was reacted, which is the same as one mole of hydrogen produced according to the chemical equation. In the end, the volume of mole of the hydrogen gas produced in this experiment would be equal to 23.6

This means that the mass of the anhydrous, AlK(SO4)2, was 0.75g and the water driven off was 0.68g. When my lab partners and I used these measurements for our calculations we got the ratio of 1:13. This ration is clearly accurate because the actual ration is

The maximum amount of product that any reaction can produce is called theoretical yield. To calculate theoretical yield, one must first write out a balanced chemical equation for the reaction to determine the number of moles of reactants used or products produced, in the reaction

What type of chemical reaction will a balloon receive when filled with an acid and base? Acids and bases are around us everywhere in various types of liquids. They are in the foods we eat and the things we use on a daily basis, for example liquid dishwashing soap. Acids are usually sour and bases are silky, giving it a bitter taste. Water can act like both, depending on the situation.

The limiting reagent in this lab was iron. Iron was the obvious limiting reactant because the 4.00 grams of iron was used to determine that 11.43 grams of copper sulfate would be necessary in the equation. Also, an extra 25% of copper sulfate was added to make sure there was enough copper sulfate in the reaction since it was the excess in the reaction. The theoretical yield of the reaction was 4.551 grams of copper. The theoretical yield is an amount predicted by stoichiometry and assumes that the limiting react is used completely; the yield was determined through stoichiometry by converting the amount of iron into the amount of copper in the reaction.

A mole is a unit of measurement that follows the rule of ; 1.00 mole = molar mass = 6.02 x 10^23 atoms/ ions/ molecules / formula units = 22.4 L of any gas at standard temperature and pressure. That formula was used to determine the mass of the anhydrated substance. A anhydrate is a substance with water heated out of it. The purpose of this lab was to determine how many moles of water are

Elements react to the different solution in different ways. Some show a chemical or physical reaction and some don’t show any kind of reaction. By using four elements and four solutions we are trying to see which out of all of these elements is the most reactive. We are trying to determine which element would be the most reactive? The elements are copper, zinc, silver, and magnesium.

In order to create 5 grams of MgSO4 from MgO (Magnesium Oxide) and H2SO4 (Sulfuric acid), we needed to create a balanced equation to find the amount of other chemicals we would need. The balanced equation was MgO + H2SO4 --> MgSO4 + H2O. After creating a balanced equation, we found the amounts of MgO and H2SO4 using stoichiometry. The amount of Magnesium Oxide was 1.674 grams and the amount of Sulfuric acid was 6.923 milliliters. In order to create as close to 5 grams of MgSO4 as possible, we decided to ignore sig-figs, and go to the hundreds place for the sake of exactness (getting an A).

Then, with the mass we calculated, we compared it with what it should have been to reach a percent yield. The percent yield we reached was 130% yield which is clearly much more than the perfect 100% yield. Oddly, the yield was greater than 100%, meaning that we ended with more copper(II) oxide than should have been possible considering how much copper(II) sulfate we started the reaction with. We indeed did have an experimental yield of 0.850 g which is almost two hundredths of a gram larger than the theoretical yield of 0.6531 g. This would be considered rather unsuccessful as a result of the almost impossible

A magnesium atom has a charge of 2+ and an oxygen atom has a charge of 2-. When the charges are balanced, the equation is MgO, one magnesium atom joined in an ionic bond with one oxygen atom. With the theoretical empirical formula found, one can now find the theoretical percent composition of MgO. The atomic molar mass of a magnesium atom is 24.31 g/mol and 16.00 g/mol for oxygen, therefore the total theoretical mass of the compound is (24.31 g/mol + 16.00 g/mol) 40.31 g/mol. The percent composition is found by dividing the element’s mass by the total mass of the

Cl_2 + 2NaOH→NaCl+NaOCl+H_2 O Initial mass (kg): 386.82 520.45 - -

Thus being said, if my partner and I were to perform this lab for a second time, we can agree that to improve our experiment results we would focus more on the chemical reaction that included zinc and hydrochloric acid, for this is the chemical reaction that we were not able to collect proper data and configure our results for. If we were able to re conduct only this part of the lab, based on previous experience, we would allow more time for the hydrogen gas to collect in hopes that adding a catalyst ( in this case heat) would evoke a reaction thus ensuring a

A higher or lower percent yield would have suggested an incorrect amount of anhydrous sodium sulfate was used or the product was not left under gentle blowing air for long enough. Percent yield was used instead of percent recovery, because the experiment involved the created of a product, not the extraction of a product

Introduction Purpose The main purpose of this experiment is to learn the principles of stoichiometry. It will help us understand the process of finding moles of each reactant, limiting reagents, and calculating theoretical and percent yields for each reaction. Theoretical Background The stoichiometry depends on the balancing chemical formulas which must have the correct ratios for each reactant, so they can form product properly.

℃^(-1)×6.40℃±3.1 %=1337.6 J±4.06 % ∆H=(-1337.6 J±4.06 %) /(0.025 mol ±0.16 %)= -53504 J m〖ol〗^(-1)±4.22 % ∆H=-53504 J m〖ol〗^(-1)±4.22 %÷1000=-54 kJ m〖ol〗^(-1)±4.22 % Conclusion and