Question

For the reaction 2AgNO3+Na2CrO4⟶Ag2CrO4+2NaNO3 how many grams of sodium chromate, Na2CrO4, are needed to react completely with 45.5 g of silver nitrate, AgNO3?

Answer 1

Answer:

28.12 g Na2CrO4

Explanation:

To solve this question, we have to work with stoichiometry:

Then, if we have 45.5 g of Silver Nitrate:

$45,5 g AgNo3 x \frac{1 mol AgNo3}{169.87 g AgNo3} = 0.2678 mol AgNo3$

We know that for each Na2CrO4's mol, we need 2 AgNo3's moles, so, we can calculate the amount of AgNo3's moles as follows:

$0.2678 mol AgNo3 * \frac{1 mol Na2CrO4}{2 Mol AgNO3} =0.134 mol Na2CrO4$

and, now we can calculate the grams of sodium chromate as:

$0.134 mol Na2CrO4 *\frac{209.9714 g Na2CrO4}{ molNa2CrO4} =28.12 g Na2CrO4$