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?

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:

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:

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