Question

1.0g of magnesium is burnt with 0.56g O2(oxygen) in a closed vessel . Which reactant is left in excess and how much ? (At. wt. Mg = 24 ; O = 16)

Answer 1

The equation is as follows:

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

Therefore 2 moles of Mg require 1 mole of O2 to produce 2 moles of MgO.

Calculate actual moles.
moles= mass/ molar mass

Mg=    1/24  = 0.041667
O2=   0.56/32 = 0.0175
MgO=

Mole ratio of Mg: O2   = 2:1

find mass of Mg that reacts:

that means 2 mole of Mg reacts with 1 moles of O2

If 1 = 2 
Then 0.0175 = 0.0175 x 2
                     =0.035 moles of magnesium

mass of magnesium that reacts:

mass= moles x molar mass
        = 24 x 0.035
         
              0.84 g
That means that of the 1 gram of magnesium that reacts only 0.84 grams is used

Therefore magnesium is in excess and by 0.16

Answer 2

Answer:

Mg , 0.16g

Explanation                                                                                                                             2Mg+ O2 =  2MgO                 

                                  48g    32g    80g -> molar mass                                                         1.0g     0.56g        -> given                                                                  To find excess reagent                                                                        (0.56 * 48)/32=0.84 g of Mg                                                                      Given mass of Mg >Needed mass of Mg                                             1.0g > 0.84     (excess of Mg available )                                 To find the amount of excess                                                             Given mass - Needed mass of Mg= 1.0 -0.84= 0.16 g