A 2.747 g sample of manganese metal is reacted with excess HCl gas to produce 3.22 L of H2(g) at 373 K and 0.9

Question

A 2.747 g sample of manganese metal is reacted with excess HCl gas to produce 3.22 L of H2(g) at 373 K and 0.951 atm and a manganese chloride compound (MnClx). What is the formula of the manganese chloride compound produced in the reaction?

Answer 1

ANSWER: MnCl4

Given the info above, you can use the Ideal gas law to find the mol of H2 used.

n = PV/RT so;

(0.951 atm)(3.22L)/(0.0821atm*L/mol*K) = 0.10mol H2

Using the mass of Mn we can find the mol of Mn:

2.747g(1mol Mn/54.938g) = 0.05 mol of Mn

As you can see, the mol ratio of Mn to H2 is 1:2.

From this you can write the balance equation between Mn and HCl to form Manganese compound. Giving 1 to 2 ratio between Mn and H2:

Mn + 4HCl ---> MnCl4 + 2H2

As you can see, the manganese compound formed is MnCl4, Manganese(IV) chloride.

Answer 2

Manganese Excess

Answer 3

scandium has mass atomic 40 5. so the mass of hydrogen produced by skill of 40 5 of Sc is so the mass of hydrogen produced is 0.1642*40 5/2.40 six=3g 3g corresponds to threeatom of hydrogen so formulation = Sc Cl3 and reaction Sc + 3 HCl---> ScCl3 +3/2 H2