You are given a sample of a Group II metal chloride, but the identity of the metal is unknown. Knowing that all Group II metal chloride compounds have the general formula, MCI2 and that the percentage of chlorine in the compound is 34.1%, what is the identity of the unknown substance?

The identity of unknown substance is Barium.

Explanation:

The given compound has a formula of MCl2. And the percentage of chlorine in this compound is said to be 34.1 %. But, we know that the mass of chlorine in this compound is, 35.5*2 = 71 g

Therefore, if we let x be the total mass of the compound, then:

34.1% of x = 71 g

(0.341)x = 71 g

x = 71 g/0.341

x = 208.2 g

Hence, the mass of other atom M, must be:

M = x - 71 g

M = 208.2 g - 71 g

M = 137.2 g

Now, we look into periodic table in group II. We find that the element is Barium with atomic mass of 137 g.

The identity of unknown substance is Barium.

i think its 5049.

i hope this !

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2