Please help with all the questions! Pure magnesium metal is often found as ribbo
ID: 1048504 • Letter: P
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Please help with all the questions!
Pure magnesium metal is often found as ribbons and can easily burn in the presence of oxygen. When 3.51 g of magnesium ribbon burns with 8.50 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.?
Pure magnesium metal is often found as ribbons and can easily burn in the presence of oxygen. When 3.31 g of magnesium ribbon burns with 8.50 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. at What is the limiting reactant? O magnesium O Oxygen The percent yield for the reaction is 84.2%, how many grams of product were recovered? Number O How many grams of the excess reactant remain? NumberExplanation / Answer
2Mg + O2 ---------> MgO
no of moles of Mg = W/G.A.Wt
= 3.51/24 = 0.146moles
no of moles of O2 = W/G.M.Wt
= 8.5/32 = 0.265moles
2moles of Mg react with 1 mole of O2
0.146moles of Mg react with = 1*0.146/2 = 0.073 moles of O2
o2 is excess reagent
Remaining excess reagent = 0.265-0.073 = 0.192moles of O2
mass of excess reagent = no of moles* gram molar mass
= 0.192*32 = 6.144g
limiting reactant is Mg
2 moles of Mg react with O2 to gives 2 moles of MgO
0.146 moles of Mg react with O2 to gives = 2*0.146/2 = 0.146 moles of MgO
mass of MgO = no of moles * gram molar mass
= 0.146*40 = 5.84g of MgO
percent yield = actual yield*100/theoretical yield
0.824 = actual Yield*100/5.84
actual yield = 0.824*5.84/100 = 0.048g
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