Prnt 11 Calculator Perodic Table stion 29 of 35 Sapling Learning m metal is ofte

Question

Prnt 11 Calculator Perodic Table stion 29 of 35 Sapling Learning m metal is often found as ribbons and can easily bum in the presence of oxygen. Pure ma When 2.92 g of magnesium ribbon burns with 7.59 g of oxygen, a bright, white light and a white, powdery product are formed Enter the balanced chemical equation for this reaction. Be sure to include all physical states Tip: If you need to clear your work and reset the equation, click the button that looks like two arrows What is the limiting reactant? O magnesium O oxygen The percent yield for the reaction is 86.0%, how many grams of product were recover ed? Number How many grams of the excess reactant remain? Number Hint Previous Give Up & View Solution 2, Check Answer O Next

Explanation / Answer

The balenced chemical equation is-

2Mg(S) + O2(g) = 2MgO

Now 7.59 g O2 = (7.59)/32 moles O2 =0.24 moles

2.92 g Mg =2.92/24 moles Mg = 0.12

Now 2 moles Mg reacts with =1 moles of O2.

0.12 moles Mg reacts with =(1*0.12)/2 moles of O2

=0.06 moles of O2.

Clearly in this reaction O2 remains excess.So the limiting reagent is magnesium.

Now clearly from the balenced equation it can be seen MgO has the same number of moles as the Mg has.So if the total Mg is exhauted then there should be 0.12 moles of MgO.But given the yield is 86% so MgO is produced = (0.12*86)/100 moles = 0.1032

Now 1 mole MgO =(24+16) = 40 g

0.1032 moles MgO = (40*.1032) = 4.28 g MgO

So 4.28 g product is recovered.

Now in 40 g MgO there is (24/40)*100 = 60% Mg and obviously rest 40% is O2.

In 4.28 g MgO there is = (4.28*60)/100 = 2.568 Mg

rest =(4.28-2.568)= 1.712 g O2.

So Mg remains excess=(2.92-2.568)= 0.352 g

O2 remains excess=(7.59-1.712)= 5.878 g O2.

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