A2.747-g sample of manganese metal is reacted with excess hcl gas to produce 3.14 l of h2(g) at 386 k and 1.01 atm and a manganese chloride compound (mnclx). what is the formula of the manganese chloride compound produced in the reaction?

The formula of the manganese chloride compound produced is MnCl4

Explanation:

Step 1:  the equation

Mn (s) +HCl (g)→  H2 (g)+ MnClx

Step 2: Calculate number of moles

p*v =n * R *T

with p= the pressure (in atm) = 1.01 atm

with v = the volume of the gas = 3.14 L

n = the number of moles

R = the gasconstant = 0.0821atm*L/mol*K

T = 386 K

1.01 * 3.14 = n * 0.0821 * 386

n = 3.1714/31.6906 = 0.1 moles H2

Step 3: Calculate moles of Mn

moles of Mn = mass of Mn / Molar mass of Mn

moles of Mn = 2.747g /54.938 g/mole = 0.05 moles of Mn

This shows the mole ratio Mn to H2 is 1:2

This gives us the equation:

Mn + HCl → 2H2 + MnClx

To balance this equation we should have 4x HCl ( to have on both sides 4H)

This gives us:

Mn + 4HCl → 2H2 + MnClx

The number of Mn is balanced on both sides

The number of Cl on the left side is 4, this means on the right side we also need 4x Cl. This means x in MnClx is 4

Mn + 4HCl → 2H2 + MnCl4

The formula of the manganese chloride compound produced is MnCl4

the answer to your question is c.

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