I would really appreciate help with this chem question.?

Question

How many grams of KCl (molar mass = 74.55 g) will be produced from the reaction of 50 mL of .3 M KOH with axcess HCl?
KOH(aq) + HCl(aq) --> KCl(aq) + H2O (l)
1. 9.31 g
2. 1.12 g
3. .876 g
4. 4.65 g
5. 6.98 g

Answer 1

To answer this question you need to know how many mols of reactants you have and to determine if one of them is limiting. You can make only as many mols of product as you have with the least number of mols of the reactants.

mols KOH = M x V where M is the concentration of KOH in mols/ l and V = the volume in l.

mols KOH = 3 m/l x 0.050 l = .150 mol KOH

Since the mols of KOH = mols of KCl then the g of KCl = MW KCl x the mols of KCl

0.15 mol KCl x 74.38 g/mol = 11.16 g

Now since there are only two significant figures in 50 ml your answer can only have two sig. figures. Therefore the correct answer would be 11g KCl

Answer 2

Moles of KOH = 0.05L x 0.3 M = 0.015 moles
moles of KCl from 1 mole of KOH=1
so moles of KCl produced in problem=0.015
Mass = 74.5 g x 0.015= (answer 2)

Answer 3

figure out how many moles of KCl you will make, then multiply by the MW

50 ml x 1liter/1000 ml x .3 moles/liter = .015 mols
.015 x 74.55 =

Answer 4

open your text book and do your homework....you will never learn if ppl give you all the answers. If you are having trouble ask a parent or a friend maybe they can help you with the equation.