Question

Calculate the number of gm of K2SO4 and water produced when 14 g of KOH are reacted with excess of H2SO4 . Also calculate the number of molecules of water produced

Answer 1

Answer:

1). number of gm of K2SO4= 25.14gm

2) 1.71×10^23 water molecules

Explanation:

°given:------------------

KOH. +. H2SO4 -----------> K2SO4 +H2O

(14gm)

°to find :---------------

•number of gramm of K2SO4

• number of water molecules produced

°solution:-------------

*balancing the equation both side :-

2KOH. +. H2SO4 -----------> K2SO4 + 2H2O

2mole 1 mole 1 mole. 2 mole

•molecular mass:-

KOH= 39+16+1=56

H2SO4= 1*2+32+64= 98

K2SO4= 176

H2O= 18

°numbers of mole present in 14 KOH

= given mass /molar mass= 16/56=4/14=2/7 mole

2 mole KOH generates 1 mole K2SO4

hence one mole KOH will generate

1/2 mole of K2SO4

then 2/7 mole KOH will generate

$\frac{2}{7} \times \frac{1}{2}$

= 1/7 moles of K2SO4

now mass of K2SO4=

$\frac{1}{7} \times 176$

=25.142857 gm

2) mass of H2O =

$\frac{2}{7} \times 18 = \frac{36}{7} gm$

18gm-------> 6.022×10^23

1gm--------->6.022×10^23/18

36/7g------>6.022×10^23 ×36/7×18=1.71×10^23 molecule