Question

Pure magnesium metal is often found as ribbons and can easily burn in the presence of oxygen. When 3.29 g of magnesium ribbon burns with 7.42 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. equation: What is the limiting reactant? O magnesium oxygen If the percent yield for the reaction is 90.7%, how many grams of product were recovered? mass of product recovered: How many grams of the excess reactant remain?

Answer 1

no of moles of Mg   = 3.29/24   = 0.137moles

no of moles of O2    = 7.42/32   = 0.231875 moles

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

1 mole of O2 react with 2 mole of Mg

0.231875 moles of O2 react with = 2*0.231875/1   = 0.46375moles of Mg is required

Mg is limiting reactant

magnesium >>>>answer

2 moles of Mg react with excess of O2 to gives 2 mole of MgO

0.137moles of Mg react with excess of O2 to gives = 2*0.137/2   = 0.137moles of MgO

mass of MgO = no of moles * gram molar mass

                       = 0.137*40 = 5.48g

Theoretical yield = 5.48g

percent yield    = actual yield *100/theoretical yield

   90.7               = x*100/5.48

90.7*5.48/100   = actual yield

4.97g           = actual yield

mas of product recover = 4,97g

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

2 moles of Mg react with 1 mole of O2

0.137 moles of Mg react with = 1*0.137/2   = 0.0685moles of O2

O2 is excess of reactant

The no of moles of excess reactant remains after complete the reaction = 0.231875-0.0685   = 0.163375moles

The amount of excess reactant remains after complete the reaction = no of moles * gram molar mass

                                                                                                            = 0.163375*32   = 5.228g>>>>answer