How many grams of LiCl are in 1 L of a 0.05 M LiCl solution

The answer is 2, 12 g of LiCl in 1 L of solution

Explanation:

We calculate the weight of 1 mol of LiCl:

Weight 1 mol LiCl= Weight Li + Weight Cl =6, 94 g + 35, 45 g= 42,39 g/mol

In a 0.05M solution it means that we have 0.05 moles of compound (in this case, LiCl) in 1 liter of solution:

1 mol42, 39 g LiCl

0,05molx=(0,05molx 42, 39 g LiCl)/ 1 mol =2, 1195 g LiCl

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g