42. When 1.254 g of sodium hydrogen carbonate (bak ing soda) is heated, it decom

Question

42. When 1.254 g of sodium hydrogen carbonate (bak ing soda) is heated, it decomposes to 0.765 g of sodium carbonate. Use the balanced chemical equation and other given information to determine the percent yield of sodium carbonate. Molar Mass of sodiaum hyudrogen carbonate 8401 gmol Molar Mass of sodim carbomate-105.99 gmo A. 48.3% D. 96.7% B. 77.0% E. 83.3% C. 99.4% Using the information provided to you on the periodie table, determine the product(s) in the following reactions 43. 3AgNO(a)NasPOdaq)- A. NR C. Ag PO: (s) +3 NaNO, (aq) B. 3AgPO ()+ NaNO, (aq) D. 3AgPO,(s) + 3 NaNO, (s) 44. HNO (a)NaOH (aq) A. NR C. NaH (aq) +N:(g)+O: (g) B. NaH (aq) +NOH (aq) + O2 (g) D. NaNO, (aq)+ HOU expect to see: A. Bright light because nitrates are highly soluble in H:O. Therefore there are many ions in solution B. Dim light because nitrates are slightly soluble in H:O. Therefore there are few ions in solution. C. No light because aluminum is not soluble in H:O. Therefore there are no ions in solution. Name the following hydrocarbons 46. CH CH CH CH A. propane C. butane B. pentane E. hexane D. I-methylpropane 47. CHCH.CH-CHCH CHy C. hexane A. pentylmethane D. 2-methylpentane B. pentane E. 4-methylpentane

Explanation / Answer

Ans 42

Moles of NaHCO3 = mass/molecular weight

= 1.254g / 84.01g/mol

= 0.014926 mol

From the stoichiometry of the reaction

2 mol NaHCO3 produces = 1 mol Na2CO3

0.014926 mol NaHCO3 produces = 0.014926/2

= 0.007463 mol Na2CO3

Mass of Na2CO3 = moles x molecular weight

= 0.007463 mol x 105.99 g/mol

= 0.7910 g = theoretical yield

% yield = actual yield x 100 / theoretical yield

= 0.765 x 100 / 0.791

= 96.71 %

Option D is the correct answer

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