Question

How many grams of sodium chromate, Na2CrO4, are needed to react completely with

56.7g of silver nitrate, AgNO3?

Answer 1

The provided information lacks the balanced chemical equation and molar masses, which prevents calculation of the mass of sodium chromate required to react with silver nitrate.

Step-by-step explanation:

The question at hand involves calculating the amount of sodium chromate (Na2CrO4) required to react completely with a given mass of silver nitrate (AgNO3). To answer this question accurately, we need to use stoichiometry, which is a section of chemistry that involves calculating the amounts of reactants and products in chemical reactions.

Unfortunately, the provided information is insufficient to directly calculate the mass of sodium chromate needed, as it lacks the balanced chemical equation for the reaction between sodium chromate and silver nitrate. Typically, this would be done by first writing and balancing the chemical equation, then converting the mass of silver nitrate to moles, using the mole ratio from the balanced equation to find the moles of sodium chromate needed, and finally converting those moles back to grams based on the molar mass of sodium chromate.

To carry out these calculations, you would also need the molar mass of the compounds involved, which is not provided here. In general, without the proper chemical equation and molar masses, it's impossible to perform the calculation requested.

Answer 2

We need 27.0 grams of Na2CrO4 to react with 56.7 grams of AgNO3

Step-by-step explanation:

Step 1: data given

Mass of silver nitrate AgNO3 = 56.7 grams

Molar mass AgNO3 = 169.87 g/mol

Molar mass of Na2CrO4 = 161.97 g/mol

Step 2: The balanced equation

2AgNO3 + Na2CrO4 → Ag2CrO4 + 2NaNO3

Step 3: Calculate moles AgNO3

Moles AgNO3 = mass AgNO3 / molar mass AgNO3

Moles AgNO3 = 56.7 grams / 169.87 g/mol

Moles AgNO3 = 0.334 moles

Step 4: Calculate moles Na2CrO4 moles needed

For 2 moles AgNO3 we need 1 mol Na2CrO4 to produce 1 mol Ag2CrO4 and 2 moles NaNO3

For 0.334 moles AgNO3 we need 0.334 / 2 = 0.167 moles Na2CrO4

Step 5: Calculate mass Na2CrO4

Mass Na2CrO4 = 0.167 moles * 161.97 g/mol

Mass Na2CrO4 = 27.0 grams

We need 27.0 grams of Na2CrO4 to react with 56.7 grams of AgNO3