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:
[tex]45,5 g AgNo3 x \frac{1 mol AgNo3}{169.87 g AgNo3} = 0.2678 mol AgNo3\\[/tex]
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:
[tex]0.2678 mol AgNo3 * \frac{1 mol Na2CrO4}{2 Mol AgNO3} =0.134 mol Na2CrO4\\[/tex]
and, now we can calculate the grams of sodium chromate as:
[tex]0.134 mol Na2CrO4 *\frac{209.9714 g Na2CrO4}{ molNa2CrO4} =28.12 g Na2CrO4[/tex]
