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Decomposition
Beginning of paper
Purpose: In this lab we will observe the products of of potassium
perchlorate (KClO4). We will then predict from our results the correct chemical
reaction equation.
Procedure: 1. Weigh out about 4.0g of KClO4 in a test tube. Record the
accurate weight below. Product Weight Before Weight After Mass of Test tube +
KClO4 41.5g 39.8g Mas .... Middle of paper
.... = .03 moles 2. The number of moles of O2 that were present in
our sample of KClO4 was .06 moles. 1.9g ¸ 32g/mole = .06 moles 3. The number of
moles of O2 lost is .02 moles. 1.7g ¸ 32g/mol = .05 moles 4. KClO4 à KCl + 202
4.0g ¸ 138.6g/mol = .03 moles ´ 202 ¸ KClO4 = .06 moles ´ 32g = 1.9g 5 Percent
Yield: 89% O2 lost 1.7g ¸ O2 Expected 1.9g
....
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Number of words: 253
Number of pages: 1 (approx. 250 words per double-spaced page)
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