1) 0.830 g of an unknown metal is reacted with excess concentrated hydrochloric

Question

1) 0.830 g of an unknown metal is reacted with excess concentrated hydrochloric acid according to the reaction below. 0.355 L of hydrogen gas was collected over water on a day when the temperature was 23.0 oC and the pressure was 1.05 atm. The vapor pressure of water at 23.0 oC is 21.1 torr. Report the partial pressure of hydrogen gas in atmospheres, the molar mass of the metal and the atomic symbol for the metal. The ideal gas law constant R = 0.0821 L*atm*K-1mol-1.

M(s) + 2 HCl(aq) MCl2(aq) + H2(g)

Need to find:

Partial pressure H2 = ......... atm

Molar mass of the metal = ............. g/mol

Atomic Symbol for the metal = ...........

I need someone to show me how to solve every question step by step, please !!! :(

Explanation / Answer

Ideal gas law: PV = nRT

So here, 1.05 * 0.355 = n * 0.0821 * (273+23)

or 0.37275 = n * 24.3016

or n = 0.015

This is the number of mole of hydrogen gas

Vapor pressure of water at same temp is 21.1 torr = 21.1 * 0.00131578947 atm = 0.0278 atm

Partial pressure = mole fraction x total pressure

= (0.015/1) * (1.05 + 0.0278)     

= 0.016 atm

Here 0.83 g metal gives 0.015 moles gas

So 1 mole gas is produced from (0.83/0.015) g metal = 55.33 g metal

The balanced chemical eqaution shows 1 mole gas is produced from 1 mole metal

So, the molar mass of the metal is 55.33 g/mole

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