What mass of a divalent metal M (atomic mass = 40) would react with excess hydrochloric acid to liberate 224cm^3 dry hydrogen gas measured at S.T.P?
8.0 g
4.0 g
0.8 g
0.4 g
Explanation
Video Explanation
No video available
Post your Contribution
Discussions (34)

First convert your cm3 to dm3 I.e
224/1000=0.224
So if 40 - 22.4dm3
X - 0.224
Cross multiply and make x the subject of the formula
X = 0.224Γ40/22.4
X = 0.4g
Thanks.
Hope this helps

The question is incomplete:Here is thr complete one
What mass of a divalent metal. M (atomic mass 40) would react with excess hydrochloric acid to liberate 22.4cm3 of dry hydrogen gas measured as s.t.p.?

What mass of a divalent metal M (atomic mass= 40)
would react with excess hydrochloric acid to liberate
22 cm3 of dry hydrogen gas measured as S.T.P?

Since it's a divalent elements x is the element
2hcl + x ------xcl2 + h2
224cm/1000 = 0.224
V/22.4= mole
0.224/22.4 =0.01mol
1mole of x. --- 1mol of h
Xmol of x. ----0.01mol
So =0.4g so the element is calcium

Mass/molar mass* molarvolume/volume.
So: let mass = m and volume=224Γ·1000=0.224
: m/40*22.4/0.224=0.4g




