a weighed amount of sodium chloride is completley dissolved in a measured volume
ID: 768322 • Letter: A
Question
a weighed amount of sodium chloride is completley dissolved in a measured volume of 4.00M ammonia solution at ice temperature, and carbon dioxide is bubbled in. Assume that sodium bicarbonate is formed until a limiting reagent is entirely used up. The solubility of sodium bicarbonate in water at ice temperature is 0.75 mol per liter. Also assume that all the sodium bicarbonate precipitated is collected and converted quantitatively to sodium carbonate. data: the mass of sodium chloride in grams is : 18.8 the volume of ammonia solution in mL is 26.9 calculate how many moles of sodium chloride are present in the original solution? How many moles of ammonia are present originally? what is the limiting reagent? a) ammonia b) sodium chloride what is the theoretical yield of sodium bicarbonate in grams?Explanation / Answer
lots of stuff going on here 1) CO2 + H2O ----> H2CO3 2) H2CO3 + NH3 ----> HCO3- + NH4+ 3) Na+ + Cl- + HCO3- ----> NaHCO3 + Cl- we could combine these into an overall equation 4) NaCl + H2O + CO2 + NH3 ----> NaHCO3 + NH4Cl we have unlimited CO2 in 1 but only 0.160 mol of NH3 in 2 so we have a max of 0.160 mol HCO3 available in 4 we have 0.251 mol NaCl and 0.160 mol NH3 ratio is 1 to 1 so NH3 LIMITS giving 0.160 mol of NaHCO3 this is in 0.0399 L so we have 0.160 mol / 0.0399 L = 4.01 M any over 0.75 M will precipitate so 4.01 - 0.75 = 3.26 mol/L will precipitate 3.26 mol/L * 0.0399 L = 0.130 mol of NaHCO3 precipitates theoretical yield is 0.160 mol * 84.02 g/mol = 13.4 g precipitated is 0.130 mol * 84.02 g/mol = 10.9 g finally when the ppt is recovered and dried by heat we get 2 NaHCO3 ----> Na2CO3 + H2O + CO2 the ratio is 2 to 1 so we get 0.065 mol of Na2CO3 0.065 mol * 106.01 g/mol = 6.89 g of Na2CO3FOLLOW THIS
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