The mass of precipitate formed if a solution containing 2.34 g of cesium hydroxide is added to a solution containing a large excess of tiso4 is 0.6387 g
Mass of CsOH = 2.34 g
Number of moles of CsOH = 149.9 g / mol
Molar mass of CsOH = Mass of CsOH / Number of moles of CsOH
= 2.34 / 149.9
Molar mass of CsOH is 0.0156 mol
2 CsOH + TiSO4 → Ti(OH)2 + CsSO4
In this reaction 2 moles of CsOH gives 1 mole of Ti(OH)2
0.0156 mol of CsOH will produce, 0.0156 / 2 = 0.0078 mol
0.0156 mol of CsOH will produce 0.0078 mol of Ti(OH)2
Mass of Ti(OH)2 = Molar mass of Ti(OH)2 / Number of moles of Ti(OH)2
= 0.0078 * 81.88
= 0.6387 g
Mass of Ti(OH)2 formed is 0.6387 g
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The mass of precipitate formed is 0.90441 g
cesium hydroxide is added to a solution containing a large excess of TiSo4
The chemical reaction is
2CsOH + TiSo4 → Ti(OH)2 + CsSo4
The above reaction the Ti(OH)2 is a precipitate formed
Molar mass = mass of a substance / mass of one mole
Mass of CsOH = 2.34 / 149.912
Mass of CsOH = 0.0156 mol
The above reaction 2 mol of cesium hydroxide yield 1 mol of titanium hydroxide
Then, 0.0156 mol of CsOH yield 0.0156 / 2 = 0.0078 mol of Ti(OH)2
The mass of Ti(OH)2 is 115.95
0.0078 mol of Ti(OH)2 = 115.95 X 0.0078
= 0.90441
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