A titration of an unknown concentration of sulfuric acid needs .0836 L of a 0.12

Question

A titration of an unknown concentration of sulfuric acid needs .0836 L of a 0.12 M lithium hydroxide solution. There is 34.3 mL of the unknown concentration sample. What is the concentration of the unknown sulfuric acid sample? 0.17 M 0.36 M 0.34 M 0.20 M If you have unlimited nitrogen available, how many grams of lithium nitride can be produced from 1.75 moles of lithium? 6Li + N_2 rightarrow 2Li_3N 18.3 g 20.3 g 58.3 g 61.0 g If you have 47.2 grams of potassium nitrate and it decomposes (see reaction below), how many moles of nitrogen are produced? 4KNO_3 rightarrow 2K_20+2N_2+5O_2 0.290 mol 0.122 mol 0.233 mol 1.73 mol

Explanation / Answer

1. H2So4 +2 LiOH -------> Li2SO4 + H2O

    H2SO4                                                   LiOH

M1 =                                                    M2 = 0.12M

V1 = 0.0343 L                                          V2   = 0.0836L

n1 = 1                                                   n2 = 2

      M1V1/n1        =      M2V2/n2

             M1        =      M2V2n1/n2V1

                          =      0.12*0.0836*1/2*0.0343 = 0.146M    >>> answer

2. 6Li + N2 ------->2 Li3N

6 moles of Li produced 2 moles of Li3N

1.75 moles of Li produced = 2*1.75/6 = 0.583 moles of Li3N

mass of Li3N = no of moles * gram molar mass of Li3N

                    = 0.583*35 = 20.405g>>.>>answer 20.3

3.4KNO3 ------> 2K2O + 2N2 + 5O2

4 moles of KNO3 decomposes to gives 2 moles of N2

4*101 g of KNO3 decomposes to gives 2 moles of N2

47.2 g of KNO3 decomposes to gives = 2*47.2/4*101   = 0.233 moles of N2 >>>> answer

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