Question

How many grams of silver chromate are produced when $250ml$ of $1.500M$ $N{a_2}Cr{O_4}$ are added to excess silver nitrate? Please balance the equation before solving the problem.

Answer 1

We need to remember that the molar mass of silver nitrate is $124g$ it has a molecular formula $AgN{O_3}$. We have to know that when silver nitrate reacts with sodium chromate it produces silver chromate and sodium nitrate. We must have to know that the molarity can be defined as the number of moles divided by the volume of the liquid in mL.

Complete answer:
We need to know that silver nitrate is an inorganic compound with chemical formula $AgN{O_3}$.This salt is a versatile precursor to many other silver compounds, such as those used in photography. It is far less sensitive to light than the halides. As we know that there are four steps involved in this stoichiometry problem: We need to write a complete balanced equation for the following:
Calculate the moles of $N{a_2}Cr{O_4}$
Use the molar ratio from the balanced equation to calculate the moles of $A{g_2}Cr{O_4}$
Calculate the mass of $A{g_2}Cr{O_4}$
Step 1.Write the balanced chemical equation.
$2AgN{O_3} + N{a_2}Cr{O_4} \to A{g_2}Cr{O_4} + 2NaN{O_3}$
Step 2.Calculate the moles of $N{a_2}Cr{O_4}$
Moles of Na2CrO4=$0.250LN{a_2}Cr{O_4} \times 1.500molN{a_2}Cr{O_4}$
$1LN{a_2}Cr{O_4} = 0.3750molN{a_2}Cr{O_4}$
Step 3.Calculate the moles of $A{g_2}Cr{O_4}$
The molar ratio from the balanced equation is
$A{g_2}Cr{O_4}:N{a_2}Cr{O_4} = 1:1$
Moles of Ag2CrO4=$0.3750molN{a_2}Cr{O_4} \times 1molA{g_2}Cr{O_4}$
$1molN{a_2}Cr{O_4} = 0.3750molA{g_2}Cr{O_4}$
Step 4.Calculate the mass of $A{g_2}Cr{O_4}$
$MassofA{g_2}Cr{O_4} = 0.3750molA{g_2}Cr{O_4} \times 331.73gA{g_2}Cr{O_4}$
$1molA{g_2}Cr{O_4} = 124gA{g_2}Cr{O_4}$

Note:
We need to know that the precipitation reaction is a double displacement reaction that involves the production of a residue of a solid form called precipitate. These reactions also occur when two or more solutions are combined with different salts resulting in the formation of insoluble salt from which the solution is precipitated. Precipitation is insoluble as water thus it settles down at the bottom.