What (minimum) volume of HCl is required to dissolve the Cd? (Cd dissolves in HCl to form Cd2+)?

Question

A sample of cadmium weighing 3.512 g is dissolved in 3.00 M HCl.

Cd(s) + 2HCl(aq) -> H2(g) + CdCl2(aq)

This is the equation. I know that we need to use ratios...but i'm still not sure where to start. It would be really nice if someone can explaine to me how to do this equation and not just GIVE me the answer. Thank you everyone for your time.

Ciao.

Answer 1

One mole Cd weighs 112g while 1 mole HCl weighs 36g. From your equation, 1 mole Cd requires 2 mole HCl. Moles Cd in your sample is 3.512/112 = 0.031242494. This requires twice as many moles of HCl = 2*0.031242494
=0.062484988
moles. Your acid concentration says 3 moles per 1000ml. Required volume is 1000ml*0.062484988
/3 =
20.82832937
ml. In short the minimum volume required is 20.83ml.

Ciao

Answer 2

Atomic weight Cd = 112.411 g/mol
Moles Cd = 3.512 g / 112.411 = 0.03124
The ratio between Cd and HCl is 1 : 2
1 : 2 = 0.03124 : x
x = moles HCl needed = 0.06248
M = moles / Volume
Volume = moles / Molarity
Volume of HCl = 0.06248 / 3.00 = 0.0208 L = 20.8 mL

Answer 3

atomic mass of Cd 112.4so 3.512g of Cd= 3.512/112.4=0.0312 moles =31.2 millimole Cd
you see that for 1 mole of Cd you need 2 moles HCl
so you need 31.2*2 =62.4 mmoles HCl
1L oh HCl = 3moles
1 mL = 3mmoles so you need 62.4/3 =20.8 mL of HCl 3M