Pure magnesium metal is often found as ribbons and can easily burn in the presen

Question

Pure magnesium metal is often found as ribbons and can easily burn in the presence of oxygen. When 4.92 g of magnesium ribbon burns with 7.29 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.

Question 13 of 13 Map Pure magnesium metal is often found as ribbons and can easily burn in the presence of oxygen When 4.92 g of magnesium ribbon burns with 7.29 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. nat is the limiting reactant? O magnesium O oxygen The percent yield for the reaction is 86.2%, how many grams of product were recovered? Number How many grams of the excess reactant remain? Number Previous Give Up & View Solution e Check Answer Next Exit Hint

Explanation / Answer

1)

we have the Balanced chemical equation as:

2 Mg + O2 ---> 2 MgO

2)

Molar mass of Mg = 24.31 g/mol

mass(Mg)= 4.92 g

we have below equation to be used:

number of mol of Mg,

n = mass of Mg/molar mass of Mg

=(4.92 g)/(24.31 g/mol)

= 0.2024 mol

Molar mass of O2 = 32 g/mol

mass(O2)= 7.29 g

we have below equation to be used:

number of mol of O2,

n = mass of O2/molar mass of O2

=(7.29 g)/(32 g/mol)

= 0.2278 mol

2 mol of Mg reacts with 1 mol of O2

for 0.2024 mol of Mg, 0.1012 mol of O2 is required

But we have 0.2278 mol of O2

so, Mg is limiting reagent

Answer: Magnesium

3)

we will use Mg in further calculation

Molar mass of MgO = 1*MM(Mg) + 1*MM(O)

= 1*24.31 + 1*16.0

= 40.31 g/mol

From balanced chemical reaction, we see that

when 2 mol of Mg reacts, 2 mol of MgO is formed

mol of MgO formed = (2/2)* moles of Mg

= (2/2)*0.2024

= 0.2024 mol

we have below equation to be used:

mass of MgO = number of mol * molar mass

= 0.2024*40.31

= 8.158 g

% yield = actual mass*100/theoretical mass

86.2= actual mass*100/8.158

actual mass=7.03 g

Answer: 7.03 g

4)

From balanced chemical reaction, we see that

when 2 mol of Mg reacts, 1 mol of O2 is formed

mol of O2 reacted = (1/2)* moles of Mg

= (1/2)*0.2024

= 0.1012 mol

mol of O2 remaining = mol initially present - mol reacted

mol of O2 remaining = 0.2278 - 0.1012

mol of O2 remaining = 0.1266 mol

Molar mass of O2 = 32 g/mol

we have below equation to be used:

mass of O2,

m = number of mol * molar mass

= 0.1266 mol * 32 g/mol

= 4.05 g

Answer: 4.05 g

Feel free to comment below if you have any doubts or if this answer do not work

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