Question

How many grams of silver chromate are produced when 250. mL of 1.500 M Na2CrO4 are added to excess silver nitrate?2AgNO₃(aq) + Na₂CrO₄(aq) → Ag₂CrO₄(s) + 2NaNO₃(aq)

Answer 1

`124.5\text{ g Silver chromate is produced}`

Step-by-step explanation:

Here, we want to get the mass of silver chromate produced

Since silver nitrate is in excess, it means that Silver chromate is the limiting reactant

Now, we need to get the number of moles of silver chromate that reacted

Mathematically:

`Number\text{ of moles = molarity }^*\text{ volume}`

From the question:

molarity is 1.5 M

volume is 250 mL (250/1000 = 0.25 L)

Thus, we have the number of moles as:

`Number\text{ of moles = 1.5 }*\text{ 0.25 = 0.375 mole}`

Looking at the equation of reaction, 1 mole of sodium chromate produced 1 mole of silver chromate

Thus., 0.375 mol of sodium chromate will produce 0.375 mol of silver chromate

Now, to get the mass of silver chromate produced, we have to multiply the number of moles by the molar mass of silver chromate

The molar mass of silver chromate is 332 g/mol

Mathematically:

`\begin{gathered} mass\text{ = number of moles }*\text{ molar mass} \\ mass\text{ = 0.375 }*\text{ 332 = 124.5 g} \end{gathered}`