Ba(NO3)2 + 2KIO3 = Ba(IO3)2 + 2k(NO3)?

Question

we have .231 Ba(NO3)2 and .929 2KIO3

how many Ba(IO3)2 do we make

everything is in grams

Answer 1

First we must calculate moles of each reactant.

0.231 g Ba(NO3)2 x (1 mole Ba(NO3)2 / 261 g Ba(NO3)2) = 0.000885 moles Ba(NO3)2

0.929 g KIO3 x (1 mole KIO3 / 214.0 g KIO3) = 0.00434 moles KIO3

Even though the balanded equation tells us that we need two moles of KIO3 per every one mole of Ba(NO3)2, we have about 50 times as much KIO3 as Ba(NO3)2. Thus, Ba(NO3)2 is the limiting reactant which will determine how much product we make.

0.000885 moles Ba(NO3)2 x (1 mole Ba(IO3)2 / 1 mole Ba(NO3)2) x (487.1 g Ba(IO3)2 =

0.431 g Ba(IO3)2

Answer 2

Find the limiting reactant, or the reactant which will run out sooner.

First, let's try Ba(NO3)2. Convert to moles by dividing by its molar mass.

.231 / 261.35 = 0.00088 moles Ba(NO3)2

Since Ba(NO3)2 and Ba(IO3)2 have a 1-1 ratio, 0.00088 moles Ba(NO3)2 is needed to react with 0.00088 moles of Ba(IO3)2.

Now, try KIO3. Convert to moles by dividing by its molar mass.

.929 / 214.001 = 0.0043 moles KIO3

1 mole of Ba(IO3)2 is yielded for every 2 moles of KIO3(according to reaction). So how many moles of Ba(IO3)2 would be yielded from 0.0043 moles KIO3?

Set up a proportion.

1/2 = x/0.0043

x = 0.0021 moles Ba(IO3)2

.231 g of Ba(NO3)2 yielded 0.00088 moles Ba(IO3)2

.929 g of KIO3 yielded 0.0021 moles Ba(IO3)2

Since Ba(NO3)2 yielded a smaller amount of Ba(IO3)2, it is the limiting reactant.

Thus, 0.00088 moles of Ba(IO3)2 will be formed. Convert to grams by multiplying by its molar mass.

0.00088 * 487.13

= 0.428 g Ba(IO3)2

Hope I helped :)