6 replies to this topic

[Marconis]

So I am reviewing for my final on Wednesday, and this question just has me baffled:

What is the limiting reactant when 31.3 g of manganese (II) chloride, 48.3 g of chlorine, and 25.7 g of water react to produce manganese (IV) oxide and how much hydrochloric acid is produced?

I determined that MnCl2 is the limiting reactant. After balancing my equation, I get:

31.3(1/126)(8/1)(36)=70 something

The answer it is marking as correct is 36.3g. What am I doing wrong? Thanks.

[hermanntrude]

I can't really check your answer without a balanced equation. I'm not too sure of the reaction here. However, the steps should go like this:

1: calculate the mass of HCl produced assuming manganese chloride is LR
2: calculate mass of HCl produced assuming chlorine is LR
3: calculate mass of HCl produced assuming water is LR
4: choose the smallest number

each of steps 1-3 can be in turn broken into smaller steps:

a) calculate number of moles of reactant by dividing mass by molar mass
b) calculate number of moles of product (HCl) by using a stoichiometric conversion factor
c) calculate mass of HCl by multiplying by molar mass

[Marconis]

MnCl2 + 3Cl2 + 4H2O ----> MnO4 + 8HCl

I am pretty sure I balanced it correctly.

[hermanntrude]

check the formula for manganese (IV) oxide. Remember the (IV) states the charge on the Mn ion, not the number of oxides.

[Marconis]

Ah, it is MnO2. Is that because the molecular formula is Mn4O2, then when you do "cris cross" it is Mn2O4, empirical formula MnO2?

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I rebalanced it and am getting the correct answer. Thanks for pointing that out! I had forgotten that the roman numeral meant the charge and not the amount.

[hermanntrude]

i didn't follow your reasoning about the "crisscross" and the molecular and empirical formula (this is an ionic compound so it doesn't have a molecular formula, because there are no molecules). The only way to reason it out is that the (IV) means the Mn is Mn4+ and you should know oxygen is almost always O2- in ionic substances, so to balance the charges you must have 2 oxides for every manganese.

[Marconis]

Ok, your reasoning is the correct one and parallels mine, so I understand. Thanks for your help