To solve this we use the
equation,
<span> M1V1 = M2V2</span>
<span> where M1 is the
concentration of the stock solution, V1 is the volume of the stock solution, M2
is the concentration of the new solution and V2 is its volume.</span>
<span>2.0 M x V1 = 0.50 M x 200 mL</span>
<span>V1 = 50 mL needed</span>
The amount of W(OH)2 needed would be 448.126 g
<h3>Stoichiometric calculation</h3>
From the equation of the reaction:
W(OH)2 + 2 HCl → WCl2 + 2 H2O
The mole ratio of W(OH)2 to HCl is 1:2
Mole of 150g HCl = 150/36.461
= 4.11 moles
Equivalent mole of W(OH)2 = 4.11/2
= 2.06 moles
Mass of 2.06 moles W(OH)2 = 2.06 x 217.855
= 448.188g
More on stoichiometric calculations can be found here: brainly.com/question/8062886
Well, you could get the mass as
ad then
, where
is the sea level weight,
the sea level accel.,
the accel. above while
the weight above.
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