Freliminary Ouestions.(Pre Lab) Answer all the questions in the space provided.

Question

Freliminary Ouestions.(Pre Lab) Answer all the questions in the space provided. 1. Balance the following equations: 5) NaO(s)+Ho)NaOH(): Ans 2NOo ii) NHy(aq)+HSo4a)NHhSO(aq) Ans.: Ans:3RO402 2. Calculate the volume of 0.2063 M HCI that would be needed to neutralize exactly 0.2505 g of MgO according to the reaction: Mgo(s) +2 HCl(a) > MgCl2(aq)+ H200) Ans.: 3. A 0.2076 g of Na COy required 20.35 mL of a hydrochloric acid solution for complete neutralization Na COj(ag)+2 HCl(aq)2 NaCI(aq)+CO(8) +H200) What is the molarity of the HCl solution? Ans.:

Explanation / Answer

1) Unbalance reactions Balance reactions

a) Na2O + H2O --------> NaOH   Na2O + H2O --------> 2NaOH

b) MgCO3 + HNO3 ----> Mg(NO3)2+CO2+H2O   MgCO3 + 2HNO3 ----> Mg(NO3)2+CO2+H2O

c) NH3 + H2SO4 ---> (NH4)2SO4 2NH3 + H2SO4 ---> (NH4)2SO4

d) H3PO4+ NaOH --> Na2HPO4+ H2O H3PO4+ 2NaOH --> Na2HPO4+ H2O

2) MgO+ 2HCl ---> MgCl2 +H2O

molar ratio of MgO and HCl is 1:2

No. of moles of MgO = wt taken/molar mass = 0.2505/40.304 = 0.00622

As the molar ratio is 1:2 so the moles of HCl = 2x0.00622 = 0.0124

Molarity of HCl = no. of moles of HCl/ vol in lts = 0.2063

0.0124/vol = 0.2063

0.0124/0.2063 = vol in lts = 0.0601 lts = 60.10 ml of HCl

3) Na2CO3 + 2HCl ---> 2NaCl + CO2 + H2O

No. of moles of sodium carbonate = wt taken/mol wt = 0.2076/105.99 = 0.00196

Molar ratio of Na2CO3 and HCl is 1:2

1 mole of Na2CO3 --- 2moles of HCl

0.00195 moles of Na2CO3 --- ? 0.00196x2 = 0.00392 moles of HCl

Molarity of HCl = No. of moles of HCl/vol in lts = 0.00392 / 0.02035 = 0.193 M of HCl

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