chemistry homework help!? (moles)?

Question

Identify the limiting reactant when 5.87g of Mg(OH)2* reacts with 12.84g of HCl to form MgCl2* and water.

* the star after the 2 means it's a subscript

please help i don't get this stuff at all!

can you write it out step by step so I can understand how exactly you figure out the problem. thanks

Answer 1

First, write a balanced equation for the reaction:

2 HCl + Mg(OH)2 --> MgCl2 + 2 H2O

Next, you'll need to do two calculations to determine the number of moles of one of the products that you could possibly form from each of your two reactants. Start by converting the mass of the reactant into moles:

5.87 g Mg(OH)2/58.3 g/mol = 0.101 moles Mg(OH)2

From the balanced equation you see that 1 mol Mg(OH)2 produces 1 mol MgCl2, multiplying your moles of Mg(OH)2 by 1 gives you the moles of MgCl2 you could possibly form:

0.101 mol Mg(OH)2 X 1 mol MgCl2/1 mol Mg(OH)2 = 0.101 mol MgCl2

Now, do the same calculations for HCl:

12.84 g HCl/ 36.45 g/mol = 0.352 mol HCl

0.352 mol HCl X 1 mol MgCl2/2 mol HCl = 0.176 mol MgCl2

Based on these calculations, you can see that the Mg(OH)2 is the limiting reactant because you can make fewer moles of your product from that reactant.

Hope this helps.

Answer 2

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Answer 3

Write down the reaction formula

Mg(OH)2 + 2HCl = MgCl2 + 2H2O

The molar ratios are 1:2 ::1 :2

Mg(OH)2
First calculate the Mr
Mg x 1 = 24.3 x 1 = 24.3
O x 2 = 15.9 x 2 = 31.98
H x 2 = 1.0 x 2 = 2.0
24.3 + 31.98 + 2.0 = 58.3 (Mr of Mg(OH)2)
Next calculate moles
moles(Mg(OH)2 = 5.87 / 58.3 = 0.101 moles

Next calculate moles of HCl
Mr ( HCl) = 36.5
Moles (HCl) = 12.84/36.5 = 0.352 moles

From molar ratios

Mg(OH)2 : HCl :: 1 : 2 :: 0.101: 0.202

As there are more moles of HCl (0.352) than are required (0.202) then the limiting reagent is Mg(OH)2.

Answer 4

Afluorine exists as molecule so 2 atom of Li reacts with one moleculeof F as - 2Li + F2 -----------------------> 2 Li F Bno . there orbits & then orbitls s.p.d.f. s has optimal 2 e p has 6e optimal d has 10 e optimal f has 14 e optimal (C) no in a collection no of shells will improve as we bypass downward '. e .g. in team II Be 4 = 1s2 ,2s2 [ 2 shells] Mg=12= 1s2,2s2,2p6,3s2 [ 3 shells ] Ca =20 = 1s2,2s22p6,3s23p6,4s2 [ 4 shells ]

Answer 5

divide the mass of the molecule by the molar mass...see if that helps any