When a certain metal interacted with hydrochloric acid, 27.75 g of chloride was formed

| September 19, 2021 | education

When a certain metal interacted with hydrochloric acid, 27.75 g of chloride was formed, and when the same mass of metal interacted with hydrobromic acid, 50 g of bromide was formed. Determine the metal.

There is not enough data, for Me – we need a mass of this substance, everything I could have calculated. Please accept, or send for revision with new data.
Given:
Metal (Me)
m (MeCl) = 27.75 g
m (MeBr) = 50 g
Identify Metal-?
Solution:
Ме + HCl = MeCl + H2
Me + HBr = MeBr + H2
1) M (MeCl) = x + 35.5
M (MeBr) = x + 80
2) Let us express v, v = m / M
v (Me) in the first is equal to v (Me) in the second, so M = m / v,
27.75 / x + 35.5 = 50 / x + 80
27.75x + 2220 = 50x + 1775
22.25x = 445
x = 20
3) M (MeCl) = 55.5 g / mol
v = 0.5 mol
v (MeCl) = v (Me) = 0.5 mol



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