If a solution containing 117.63 g of silver chlorate is allowed to react completely with a solution containing 10.23 g of lithium bromide, how many grams of solid precipitate will be formed

18.8 g

Explanation:

The equation of the reaction is;

AgClO3(aq) + LiBr(aq)>LiClO3(aq) + AgBr(s)

Number of moles of AgClO3 = 117.63 g/191.32 g/mol = 0.6 moles

Number of moles of LiBr = 10.23 g/86.845 g/mol  = 0.1 moles

Since the molar ratio is 1:1, LiBr is the limiting reactant

Molar mass of solid AgBr = 187.77 g/mol

Mass of precipitate formed = 0.1 moles * 187.77 g/mol

Mass of precipitate formed = 18.8 g

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.