Question

question 8
how many grams of solid product will be produced from 17.09 g of cobalt (iii) oxide and 7.82 g of solid carbon? round your answer
to two decimal places and do not include units. 2 co₂o₃ (s) + 3 c (s) → 4 co (s) + 3 co₂ (g)
add your answer
1 point

Explanation:

Step1: Calculate molar masses

Molar mass of \(Co_2O_3\): \(2\times58.93 + 3\times16.00= 165.86\space g/mol\)
Molar mass of \(Co\): \(58.93\space g/mol\)
Molar mass of \(C\): \(12.01\space g/mol\)
Molar mass of \(CO_2\): \(12.01+2\times16.00 = 44.01\space g/mol\)

Step2: Find moles of reactants

Moles of \(Co_2O_3\): \(n_{Co_2O_3}=\frac{17.09\space g}{165.86\space g/mol}\approx0.103\space mol\)
Moles of \(C\): \(n_{C}=\frac{7.82\space g}{12.01\space g/mol}\approx0.651\space mol\)

Step3: Determine limiting reactant

From the balanced equation \(2Co_2O_3(s)+3C(s)\to4Co(s)+3CO_2(g)\), the mole ratio of \(Co_2O_3\) to \(C\) is \(2:3\).
For \(0.103\space mol\) of \(Co_2O_3\), moles of \(C\) required \(n_{C_{required}}=\frac{3}{2}\times0.103 = 0.1545\space mol\)
Since \(0.1545\space mol<0.651\space mol\), \(Co_2O_3\) is the limiting reactant.

Step4: Calculate moles of product (Co)

From the balanced equation, mole ratio of \(Co_2O_3\) to \(Co\) is \(2:4 = 1:2\)
Moles of \(Co\) produced \(n_{Co}=2\times n_{Co_2O_3}=2\times0.103 = 0.206\space mol\)

Step5: Calculate mass of Co

Mass of \(Co\): \(m_{Co}=n_{Co}\times58.93\space g/mol=0.206\times58.93\approx12.14\space g\)

Answer:

\(12.14\)