A weighed amount of sodium chloride is completely dissolved in a measured volume

Question

A weighed amount of sodium chloride is completely dissolved in a measured volume of 4.00 M ammonia solution at ice temperature, and carbon dioxide is bubbled in. Assume that sodium bicarbonate is formed until the 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 The mass of sodium chloride in (g) is 14.3 The volume of ammonia solution in (mL) is 29.8 Calculate the following How many grams of the sodium bicarbonate formed will precipitate from the above volume of solution? How many grams of sodium carbonate will be produced from this precipitate??

Explanation / Answer

Ok, so I'm assuming that the chemical formula is written as - 3H2 + N2 ----> 2NH3 2.80 = moles of N2 17.03052 g/mol = Molar mass of NH3 (2.80 mol N2) x (2 NH3) / (1 N2) = 5.6 mol NH3 x (17.03052 g) / (1 mol NH3) = 95.4 g NH3 a) ammonia Ca(NO3)2 + Na2CO3 = CaCO3 + 2NaNO3 Yes a precipitate of Calcium Carbonate is formed since it is insoluble in water. Mol Wt of Calcium Nitrate is 164. And that of Calcium Carbonate is 100. One mole of Calcium Nitrate produces one mole of Calcium Carbonate. i.e. 164 gms will produce 100gms of precipitate So, 1.74gms of Calcium Carbonate will be obtained from 2.85gms Calcium Nitrate present in the original solution.

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