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
Pure magnesium metal is often found as ribbons and can easily burn in the presence of...
Pure magnesium metal is often found as ribbons and can easily burn in the presence of oxygen. When 4.25 g of magnesium ribbon burns with 8.15 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: 2Mg(s) +02(g) 2MgO(s) What is the limiting reactant? охуgen magnesium If the percent yield for the reaction is 82.2%, how many grams of product...
Pure magnesium metal is often found as ribbons and can easily burn in the presence of oxygen. When 4.20 g of magnesium ribbon burns with 6.75 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? oxygen OR magnesium If the percent yield for the reaction is 79.9%, how many grams of product were recovered?...
Pure magnesium metal is often found as ribbons and can easily burn in the presence of oxygen. When 3.19 g of magnesium ribbon burns with 6.71 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: 2Mg +0, 2MgO / / What is the limiting reactant? O magnesium O oxygen The reaction goes to completion, but in...
Pure magnesium metal is often found as ribbons and can easily burn in the presence of oxygen. When 2.83 g magnesium ribbon burns with 8.82 g oxygen, 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.
I’m not sure what I’ve done wrong since the rest of my calculations were correct, so will someone please explain to me how to solve these. I’ve tried everything i can think of and nothing is working. We were unable to transcribe this imagebalanced chemical equation: Pb(NO), (aq) + 2Cl(aq) - PbCl2(s) + 2KNO3(aq) What is the limiting reactant? lead(II) nitrate O potassium chloride The percent yield for the reaction is 80.6%. How many grams of precipitate is recovered? precipitate...
Magnesium oxide can be made by heating magnesium metal in the presence of oxygen. The balanced equation for the reaction is: 2Mg(s)+O2(g)→2MgO(s) When 10.1 g of Mg are allowed to react with 10.5 g of O2, 13.1 g of MgO are collected. -Determine the limiting reactant for the reaction. -Determine the theoretical yield for the reaction. -Determine percent yield for the reaction.
Magnesium oxide can be made by heating magnesium metal in the presence of oxygen. The balanced equation for the reaction is: 2Mg(s)+O2(g)→2MgO(s) When 10.1 g of Mg are allowed to react with 10.5 g of O2, 13.9 g of MgO are collected. a) Determine the limiting reactant for the reaction. b) Determine the theoretical yield for the reaction. c) Determine percent yield for the reaction.
ReviewI ConstantsI Periodic Table Part A Magnesium oxide can be made by heating magnesium metal in the presence of the oxygen The balanced equation for the reaction is Determine the limiting reactant for the reaction. 2 Mg(s) +O2(g) 2 MgO(s) Mg(s) 02 (g) When 10.2 g Mg is allowed to react with 10.4 g O2, 12.0 g MgO is collected You may want to reference (Pages 146-151) section 4.3 while completing this problem. Previous Answer Correct The limiting reactant is...
Magnesium oxide can be made by heating magnesium metal in the presence of the oxygen. The balanced equation for the reaction is 2 Mg(s)+O2(g) 2 MgO(s) Consider that you react 12.62 g Mg with 13.08 g O2 gas. What is the theoretical yield of MgO that can be generated from this reaction? Enter a numerical answer only to three significant figures, in terms of grams.
An aqueous solution containing 7.42 g of lead(II) nitrate is added to an aqueous solution containing 6.44 g of potassium chloride. Enter the balanced chemical equation for this reaction. Be sure to include all physical states. balanced chemical cquation: What is the limiting reactant? lead(II) nitrate potassium chloride The percent yield for the reaction is 89.3 %. How many grams of precipitate is recovered? precipitate recovered: How many grams of the excess reactant remain? The percent yield for the reaction...