Question

question 13 of 24 > pure magnesium metal is often found as ribbons and can easily burn in the presence of oxygen. when 3.00 g of magnesium ribbon burns with 8.01 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) + o₂(g) → 2mgo(s) what is the limiting reactant? magnesium oxygen if the percent yield for the reaction is 84.1%, how many grams of product were formed? mass of product formed: 4.18 g how many grams of the excess reactant remain? mass of excess reactant: g tools x10^y

Explanation:

Step1: Calculate moles of reactants

Molar mass of Mg is approximately 24.31 g/mol, so moles of Mg = $\frac{3.00\ g}{24.31\ g/mol}= 0.123\ mol$. Molar mass of $O_2$ is approximately 32.00 g/mol, so moles of $O_2=\frac{8.01\ g}{32.00\ g/mol}=0.250\ mol$.

Step2: Determine mole - ratio from balanced equation

From the balanced equation $2Mg(s)+O_2(g)\to2MgO(s)$, the mole - ratio of Mg to $O_2$ is 2:1.

Step3: Identify limiting reactant

For 0.250 mol of $O_2$, the amount of Mg required is $2\times0.250 = 0.500$ mol. But we have only 0.123 mol of Mg. So Mg is the limiting reactant.

Step4: Calculate theoretical yield of MgO

From the balanced equation, 2 moles of Mg produce 2 moles of MgO. So 0.123 mol of Mg will produce 0.123 mol of MgO. Molar mass of MgO is approximately $24.31 + 16.00=40.31$ g/mol. Theoretical yield of MgO = $0.123\ mol\times40.31\ g/mol = 4.96\ g$.

Step5: Calculate actual yield

Percent yield = $\frac{\text{Actual yield}}{\text{Theoretical yield}}\times100\%$. Given percent yield = 84.1%, so actual yield = $\frac{84.1}{100}\times4.96\ g = 4.18\ g$.

Step6: Calculate moles of $O_2$ reacted

From the balanced equation, 2 moles of Mg react with 1 mole of $O_2$. So moles of $O_2$ reacted with 0.123 mol of Mg is $\frac{0.123}{2}=0.0615$ mol.

Step7: Calculate moles of $O_2$ remaining

Initial moles of $O_2$ is 0.250 mol and moles of $O_2$ reacted is 0.0615 mol. So moles of $O_2$ remaining = $0.250 - 0.0615=0.1885$ mol.

Step8: Calculate mass of $O_2$ remaining

Mass of $O_2$ remaining = $0.1885\ mol\times32.00\ g/mol = 6.03\ g$.

Answer:

6.03