Penocic Table -Print Calculator 23 of 26 Sapling Learning A sample of 5.22 g of

Question

Penocic Table -Print Calculator 23 of 26 Sapling Learning A sample of 5.22 g of solid calcium hydroxide is added to 38.5 mL of 0.170 M aqueous hydrochloric acid Enter the balanced chemical equation for the reaction. Physical states are optional and not graded. Tip: If you need to clear your work and reset the equation, click the button that looks like two arrows. What is the limiting reactant? O calcium hydroxide O hydrochloric acid How many grams of salt are formed after the reaction is complete? Number How many grams of the excess reactant remain after the reaction is complete? Number O previous Q) Check Answer NextExit

Explanation / Answer

i)

Ca(OH)2(s) + 2HCl(aq) - - - - - > CaCl2(aq) + 2H2O(l)

ii)

No of mole of HCl = (0.170mol/38.5ml)×1000ml = 0.006545

No of mole of Ca(OH)2 = 5.22g/74.093g/mol =0.07045mole

stoichiometrically, 1mole of Ca(OH)2 require 2mole of HCl. So, 0.07045mole of Ca(OH)2 require 0.1409 mole of HCl, but availble mole of HCl is 0.006545

Therefore,

Limiting reagent is HCl

iii)

Stoichiometrically, 2mole of HCl produce 1mole of CaCl2

No of mole of CaCl2 produced = 0.006545mole/2=0.0032725mol

Molar mass of CaCl2 = 110.98g/mol

mass of CaCl2 produced = 110.98g/mol ×0.0032725mol =0.3632g

iv)

Stoichiometrically, 2mole of HCl react with 1mole of Ca(OH)2

0.006545mole of HCl react with 0.0032725mole of Ca(OH)2

No of mole of Ca(OH)2 remaining = 0.07045mol - 0.0032725 = 0.067178mol

molar mass of Ca(OH)2 = 74.093g/mol

mass of Ca(OH)2 remaining = 74.093g/mol × 0.067178mol =4.9774g

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