Equations: Na OH(aq) + HCl(aq) rightarrow NaCl(aq) + H2O(l) used rightarrow Na2C

Question

Equations: Na OH(aq) + HCl(aq) rightarrow NaCl(aq) + H2O(l) used rightarrow Na2CO(aq) +HCl rightarrow NaCl(aq) + CO2 + H2O(l) Na2O(aq) + HCl(aq) rightarrow NaCl(aq) + H2O(g) Masss of Crucible 19.00 Initial mass of crucible and baking soda 24.00 Final mass of crucible and solid product 23.70 Mass of solid 1 y concentration of HCl 1 M Volume of HCl to reach the end point 26.3 ML Mass of Evlemeyer Flask Mass of Eveumeyer Flask and solid 144.5 Mass of Evlenmeyer Flask and Nacl 0.384 Stoichiometric Analysis You heated 5.00 g of baking soda in a crucible to produce 4.70 g of a solid product. The product could be NaOH, Na2CO3, or Na2O. What mass of NaOH could you have produced? This value is not correct. Please check your calculations and try again. What mass of Na2CO3 could you have produced? What mass of Na2O could you have produced? We can evaluate the three possible products by comparing the percent errors 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 Na?0, what is the percent error? Using your results, which of the following equations best describes the thermal decomposition of baking soda? Incorrect: NaHCO3(s) rightarrow NaOH(s) + CO2(g) correct: 2NaHCO3(s) rightarrow Na2CO3(s) + cO2(g) +- H2O(g) Incorrect: 2NaHCO3(s) rightarrow Na2O(s) + 2CO2(g) + H2O(g) Incorrect: With these data, at least two of the above are possible. You are correct. Your receipt no. is 151-1534

Explanation / Answer

no. of mol of NaHCO3 = 5/84.007 =0.0595 mol

NaHCO3------>NaOH + CO2

so, wt of NaOH =0.0595*39.997 =2.379 g

2NaHCO3 ? Na2CO3 + CO2 + H2O

so, 2 mol of 2NaHCO3 produces 1 mol of Na2CO3

so no. of mol ofNa2CO3 produced =0.0595/2 =0.02975 mol

wo wt ofNa2CO3 produced =0.02975*105.9784 =3.152 g

Na2CO3 (+heat) ---> Na2O + CO2

0.02975 mol ofNa2CO3 will produce0.02975 mol of Na2O

so, wt of Na2O =0.02975*61.9789=1.843 g

so ans are

1)2.379gm . no NaOH will be produced

2)wt ofNa2CO3 could have produced =3.152 g

3)wt of Na2Ocould have produced=1.843 g

4)% error for NaOH = 100-(2.379/4.70)*100=49.382 %

5) % error for Na2CO3 =100-( 3.152/4.70)*100 =32.937 %

6)% error for Na2O =100-( 1.843/4.70)*100 =60.788%

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