Question

Given :

2HCl(aq) + Mg(s) -> MgCl2(aq) +H2(g)

Mg(s) + 1/2 O2(g) -> MgO(s)

Mass of 3.00M HCl (diluted in water) : 88.4g

Vol of 3.00M HCl : 87.5mL

Mass of Mg : 0.599g

Total mass: 88.999g

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Find mole of product IF: (show calculations)

A. HCl is the limiting reactant: ______

B. Mg is the limiting reactant: ______

Answer 1

2HCl(aq) + Mg(s) -> MgCl2(aq) +H2(g)

from the balanced chemical equation,

2 moles of HCl is reacting with 1 mole of Mg and forming---- 1 mole of MgCl2 and 1mole of H2(g)

that is 73g of HCl is reacting with 24g of Mg and forming----- 95g of MgCl2 and 2g of H2.

If HCl is the limiting reagent,then

88.4g HCl gives--------------------------? g of MgCl2

= 88.4g x 95g / 73g

= 115.0g of MgCl2 = 1.21 moles                                since number of moles= given mass/molecular mass

88.4g HCl gives--------------------------? g of H2

= 88.4g x 2g / 73g

= 2.42g of H2.   = 1.21 moles

If Mg is the limiting reagent: then

24g of Mg is giving-------- 95g of MgCl2

then 0.599g of Mg gives----?

= 0.599g x 95g/24g

=2.37g of MgCl2 = 0.025moles

24g of Mg is giving-------- 2g of H2

then 0.599g of Mg gives----?

= 0.599g x 2g/24g

=0.0499

=0.05g of H2= 0.025moles

In the second reaction: if Mg is the limiting reagent: then

Mg(s) + 1/2 O2(g) -> MgO(s)

from the balanced chemical equation 1mole of Mg is reacting with 1/2 mole of O2 and forming---- 1mole of MgO

that is 24g of Mg is reacting with 16g of O2 and forming ------ 40g of MgO

If Mg is the limiting reagent, then

o.599g of Mg forms how many grams of MgO

= 0.599g x 40g/24g

= 0.998g of MgO = 0.025moles