Question

question 8
how many grams of solid product will be produced from 19.04 g of cobalt (iii) oxide and 6.88 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
page 8 of 8
1 point

Explanation:

Step1: Calculate moles of reactants

Molar mass of \(Co_2O_3\): \(2\times58.93 + 3\times16= 165.86g/mol\)
Moles of \(Co_2O_3=\frac{19.04g}{165.86g/mol}\approx0.115mol\)
Molar mass of \(C\): \(12.01g/mol\)
Moles of \(C=\frac{6.88g}{12.01g/mol}\approx0.573mol\)

Step2: Determine limiting reactant

From the balanced equation \(2Co_2O_3(s)+3C(s)\to4Co(s)+3CO_2(g)\), mole ratio of \(Co_2O_3:C = 2:3\)
For \(0.115mol\) of \(Co_2O_3\), moles of \(C\) required \(=\frac{3}{2}\times0.115mol = 0.1725mol\)
Since \(0.573mol>0.1725mol\), \(Co_2O_3\) is the limiting reactant.

Step3: Calculate moles of \(Co\) produced

From the balanced equation, mole ratio of \(Co_2O_3:Co = 2:4 = 1:2\)
Moles of \(Co = 2\times0.115mol=0.23mol\)

Step4: Calculate mass of \(Co\)

Molar mass of \(Co = 58.93g/mol\)
Mass of \(Co=0.23mol\times58.93g/mol\approx13.55g\)

Answer:

13.55