What volume of 0.125 M cobalt(III) sulfate is required to react completely with

Question

What volume of 0.125 M cobalt(III) sulfate is required to react completely with 4.00 g of potassium carbonate if the reaction proceeds according to the following equation?

Co2(SO4)3(aq) + 3 K2CO3(s) -> Co2(CO3)3(s) +3 K2SO4(aq)  The volume is_____x10^____L.  EXPLAIN

What volume of 0.125 M cobalt(III) sulfate is required to react completely with 4.00 g of potassium carbonate if the reaction proceeds according to the following equation? Co2(SO4)3(aq) + 3 K2CO3(s) -> Co2(CO3)3(s) +3 K2SO4(aq) The volume is x10^ L. EXPLAIN

Explanation / Answer

First you need a balanced reaction
Co2(SO4)3 + 3 Na2CO3 -----------> Co2(CO3)3 + 3 Na2SO4
So, you need 1 moles of cobalt (III) sulfate for every 3 moles of sodium carbonate.
Change 5 grams of sodium carbonate to moles by dividing by the molar mass.
5 g Na2CO3/(106g/mole)
Divide that by 3 to get the number of moles of cobalt(III) sulfate needed.
But the cobalt (III) sulfate is divided in water, so take the number of moles of cobalt (III) sulfate you just solved for and divide by .2500 moles per liter to get the volume needed in liters. I'd convert to milliliters by multiplying by 1000, but you wouldn't need to.

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