In the equation $Mg + 2HCl -> MgCl_2 + H_2$ , what volume of hydrogen at STP is produced from the reaction of 50.0 g of $Mg$ and the equivalent of 75 g of $HCl$ ?

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Answer 1 · 2016-02-28T06:54:56

$1.03$ mol of dihydrogen gas will evolve, with a volume slightly over $22.4$ $dm^3$ at $STP$ .

Moles of magnesium: $(50.0*g)/(24.31*g*mol^-1)$ $=$ $2.06$ $mol$

Moles of hydrogen chloride gas: $(75.0*g)/(36.2*g*mol^-1)$ $=$ $2.07$ $mol$

Given equimolar amounts, clearly, the gas is in deficiency as the stoichiometric equation requires 2 equiv of gas per equiv of metal.

So, the $1$ mole of gas evolves, (conveniently, as at $STP$ the molar volume is $22.4$ $dm^3$ ).

In the equation Mg + 2HCl -> MgCl_2 + H_2Mg + 2HCl -> MgCl_2 + H_2, what volume of hydrogen at STP is produced from the reaction of 50.0 g of MgMg and the equivalent of 75 g of HClHCl?