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  1. Problems
  2. /Chromite ore is roasted with sodium carbonate and air: FeCr₂O₄ + Na₂CO₃ + O₂ → Na₂CrO₄ + Fe₂O₃ +…
MediumChemistryHigh School (11–12)The d- and f-Block Elements › Transition Elements (d-Block)Numerical

Chromite ore is roasted with sodium carbonate and air: FeCr₂O₄ + Na₂CO₃ + O₂ → Na₂CrO₄ + Fe₂O₃ +…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

Chromite ore is roasted with sodium carbonate and air: FeCr2O4+Na2CO3+O2→Na2CrO4+Fe2O3+CO2\text{FeCr}_2\text{O}_4 + \text{Na}_2\text{CO}_3 + \text{O}_2 \rightarrow \text{Na}_2\text{CrO}_4 + \text{Fe}_2\text{O}_3 + \text{CO}_2FeCr2​O4​+Na2​CO3​+O2​→Na2​CrO4​+Fe2​O3​+CO2​. Find the electrons lost by one formula unit of chromite and the electrons gained by one O2\text{O}_2O2​ molecule. Then balance the equation for 4 formula units of chromite, and give the coefficients of O2\text{O}_2O2​ and of Na2CO3\text{Na}_2\text{CO}_3Na2​CO3​.

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Need a reminder? The key ideaShowHide

Count the electrons lost and gained from the oxidation numbers. Match them with the smallest numbers that both divide into. Then balance the other atoms by counting.

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Step 1. Electrons lost. Iron rises by 1. Two chromium atoms each rise by 3, which is 6. One formula unit of chromite loses 1+6=71 + 6 = 71+6=7 electrons.

Step 2. Electrons gained. Two oxygen atoms each fall by 2, so one O2\text{O}_2O2​ gains 4 electrons.

Step 3. Match. The smallest number that 7 and 4 divide into is 28. So 28÷7=428 \div 7 = 428÷7=4 formula units of chromite and 28÷4=728 \div 4 = 728÷4=7 molecules of oxygen.

Step 4. The rest. Four chromite units hold 8 chromium atoms, so 8 Na2CrO4\text{Na}_2\text{CrO}_4Na2​CrO4​. Each has 2 sodium atoms, so 16 sodium atoms, which means 8 Na2CO3\text{Na}_2\text{CO}_3Na2​CO3​. Four iron atoms give 2 Fe2O3\text{Fe}_2\text{O}_3Fe2​O3​, and 8 carbon atoms give 8 CO2\text{CO}_2CO2​.

Check. Oxygen: 16+24+14=5416 + 24 + 14 = 5416+24+14=54 on the left. On the right, 32+6+16=5432 + 6 + 16 = 5432+6+16=54.

Answer: 4FeCr2O4+8Na2CO3+7O2→8Na2CrO4+2Fe2O3+8CO24\text{FeCr}_2\text{O}_4 + 8\text{Na}_2\text{CO}_3 + 7\text{O}_2 \rightarrow 8\text{Na}_2\text{CrO}_4 + 2\text{Fe}_2\text{O}_3 + 8\text{CO}_24FeCr2​O4​+8Na2​CO3​+7O2​→8Na2​CrO4​+2Fe2​O3​+8CO2​. The answers are 7, 4, 7 and 8.

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