Iodine is prepared both in the laboratory and commercially by adding cl2(g)cl2(g) to an aqueous solution containing sodium iodide. 2nai(aq)+cl2(g)⟶i2(s)+2nacl(aq) 2nai(aq)+cl2(g)⟶i2(s)+2nacl(aq) how many grams of sodium iodide, nai, nai, must be used to produce 89.1 g89.1 g of iodine, i2? i2? mass: g nai

Answer : The mass of sodium iodide used to produced must be, 105.22 grams.

Explanation : Given,

Mass of = 89.1 g

Molar mass of = 253.8 g/mole

Molar mass of = 149.89 g/mole

First we have to calculate the moles of .

Now we have to calculate the moles of .

The balanced chemical reaction is,

From the balanced chemical reaction, we conclude that

As, 1 mole of obtained from 2 moles of

So, 0.351 moles of obtained from moles of

Now we have to calculate the mass of .

Therefore, the mass of sodium iodide used to produced must be, 105.22 grams.

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !