NaOH(aq) + HCl(aq) rightarrow NaCl(aq) + H2O(l) Na2O(aq) + HCl(aq) rightarrow Na
ID: 771213 • Letter: N
Question
NaOH(aq) + HCl(aq) rightarrow NaCl(aq) + H2O(l) Na2O(aq) + HCl(aq) rightarrow NaCl(aq) + CO2(g) + H2O(l) Na2O(aq) + HCl(aq) rightarrow NaCl(aq) + H2O(q) Mass of crucible 19.00 Initial mass of crucible and baking god a 21.00 Final mass of crucible and solid 1g concentration of HCl 1M Volume of HCl to rach the and point 26.3ML Mass of Evlenmeyer flask Mass of Evlenmeyer falsk and solid 14.45 Mass of Evleumeyer of flask and Nacl 0.38g Stoichiometric Analysis You heates 5.00 g of baking soda in a crubile to produce 4.70 g of a solid product. The product could be NaOH, Na2O3, or Na2O. What mass of NaOH could you have produced? What mass of Na2CO3 could you have produces? What mass of Na2O Could you have produced? We can evalute the three possible products by comparing the percent error of the mass produced. You produced 4.70 g of product. We will compare the magnitude (absolute value) of the percent errors, so do not worry about the sign. Assuming you produced NaOH,what is the percent error? Assuming you produced Na2CO3,what is the percent error? Assuming you produced Na2O,what is the percent error? Using your results, which of the following equations best describes the thormal docomposition of baking soda? NaHCO2(s) rightarrow NaOH(s) + CO2(g) 2NaHCO3(s) rightarrow Na2CO3(s) + CO2(g) + H2O(g) 2NaHCO3(s) rightarrow Na2O (s) + 2CO(s) + 2CO2(g) + H2O(g) With these data, at least two of the above are possible. According to your data from all three parts of this lab, which of the following equations describes the thermal decomposition of baking soda? NaHCO2(s) rightarrow NaOH(s) + CO2(g) 2NaHCO3(s) rightarrow Na2CO3(s) + CO2(g) + H2O(g) 2NaHCO3(s) rightarrow Na2O (s) + 2CO(s) + 2CO2(g) + H2O(g) With these data, at least two of the above are possible.Explanation / Answer
http://answers.yahoo.com/question/index?qid=20080121202642AAkvaQg http://forum.unity3d.com/threads/120357-Unity-NaCl-question u will get idea by this
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