Hugh Kim Lab Report: Stoichiometry Lab 1. Prelab Part1. 1) Create no waste = The principle that encourages chemists to not create waste at the first place rather than cleaning it up afterwards effectively shifts the chemistry more environmentally conscious‚ as creating no waste would make the experiment efficient; the reactants will be reduced to only the essential ones and the product will be maximized‚ a change that would make the experiment economic. Also‚ if chemists aim to
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Na2CO3(s)+CO2(g)+H2O(g). The was the only equation that matched up exactly with my data in terms of percentage. To start off with‚ when I balanced out the equation‚ I got 2 NaHCO3→ 1 Na2CO3(s)+ 1 CO2(g)+ 1 H2O(g). Therefore when I set up my stoichiometry problem I got 3.2 grams NaHCO3 over 1 x 1 mol NaHCO3 over 84.007g NaHCO3 x 1mol Na2CO3 over 2 mol NaHCO3 x 105.987g Na2CO3 x 1 mol Na2CO3. Hence‚ I multiplied 3.2 x 1 x 1 x 105.987 and got 339.1584. Afterwards‚ I divided 339.1584 by 84.007 and
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IB CHEMISTRY Stoichiometry Lab Data Collection and Processing Item | Mass | Small beaker (100 mL) | 47.0 grams | Large beaker (150 mL) | 82.4 grams | Mass of filter paper | 0.50 grams | Mass of coffee filter | 1.00 gram | 150mL beaker + 20mL water + lead nitrate solution | 96.1 grams | 100mL beaker + 20mL water + sodium carbonate solution | 64.2 grams | Watch glass | 32.2 grams | Precipitate + filter paper + coffee filter | 2.20 grams | Precipitate + 150mL + coffee filter
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Stoichiometry Practice Problems Be sure to balance all equations before you start calculating! Answer key is at the end. 1) Lithium hydroxide reacts with hydrobromic acid to produce lithium bromide and water. If you start with ten grams of lithium hydroxide‚ how many grams of lithium bromide will be produced? 2) Ethylene (C2H4 ) reacts with oxygen gas to produce carbon dioxide and water. If you start with 45 grams of ethylene‚ how many grams of carbon dioxide will be produced?
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Stoichiometry II- Synthesis of a Coordination Compound Report 〖CuSO〗_4 (s)→〖Cu〗^(2+) (aq)+ SO_4 (aq)^(2-) Cu^(2+) (aq)+ 4NH_3 (aq)→[Cu(NH_3 )_4 ]^(2+) [Cu(NH_3 )_4 ]^(2+)+ SO_4 (aq)^(2-)+H_2 O□(→┴(methanol solvent) ) [Cu(NH_3 )_4 ]SO_4*H_2 O 1.992 g 〖CuSO〗_4 (s) × (1 mol〖 CuSO〗_4 )/(159.6 g) = .0124 mol 〖CuSO〗_4 (s) .0124 mol 〖CuSO〗_4 (s)×(1 mol [Cu(NH_3 )_4 ]SO_4*H_2 O )/(1 mol〖 CuSO〗_4 )×(245.74 g [Cu(NH_3 )_4 ]SO_4*H_2 O )/(1 mol [Cu(NH_3 )_4 ]SO_4*H_2 O )=3.047 g [Cu(NH_3 )_4 ]SO_4*H_2
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Basic Stoichiometry PhET Lab rvsd 2/2011 Let’s make some sandviches! _ Introduction: When we bake/cook something‚ we use a specific amount of each ingredient. Imagine if you made a batch of cookies and used way too many eggs‚ or not enough sugar. YUCK! In chemistry‚ reactions proceed with very specific recipes. The study of these recipes is stoichiometry. When the reactants are present in the correct amounts‚ the reaction will produce products. What happens if there are more or less of
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Copper-Iron Stoichiometry Lab Report 10/3/12 Abstract: The lab performed required the use of quantitative and analytical analysis along with limiting reagent analysis. The reaction of Copper (II) Sulfate‚ CuSO4‚ mass of 7.0015g with 2.0095g Fe or iron powder produced a solid precipitate of copper while the solution remained the blue color. Through this the appropriate reaction had to be determined out of the two possibilities. Through the use of a vacuum filtration system the mass of Cu was
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For those of you who came to the review in class today‚ The stoichiometry question from class. I read the answer to the wrong decimal place. --What is the total mass of oxygen in 0.26mol of Cu2O? Find the percent of oxygen in Cu2O first. The molar mass of Cu2O is 143‚ the mass of just the oxygen in it is 16. % Composition of oxygen would be 16/143 times 100 is 11.19%‚ To determine the mass of Cu2O in 0.26 mol‚ multiply by the molar mass of Cu2O. This gives
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precipitation reaction. It is also to learn concepts of solubility and the formation of a precipitate. The Experiment and Observation: Weigh out your 1.0g of CaCl2-2H20 and put it into the 100mL beaker‚ add your 25mL of distilled water and stir to form the calcium chloride solution. Next‚ use stoichiometry to determine how much Na2CO3 and put it into a small paper cup. Then add the 25mL of distilled water to make the sodium carbonate solution. Mix the two solutions in the beaker and a precipitate of calcium
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Ch. 1 Fundamentals of Chemistry Subchapter – Stoichiometry and Units 1. An element X forms an oxide with the formula X4O10. If 31 g of X combines with 40 g of oxygen‚ what is the identity of X? A. N B. P C. C D. S Correct Answer: B: P Explanation: We can solve for the atomic mass of X to identify it. The amount in g of X for any amount in g of O is given by: Rearranging the above equation and substituting the given mass of X and O‚ we get: Thus‚ the atomic mass of X
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