We wish to determine how many grams of [tex]$KNO _3$[/tex] can form when 100 mL of [tex]$0.40 \, M \, K _2 CrO _4$[/tex] solution is added to excess [tex][tex]$AgNO _3$[/tex][/tex].

The balanced chemical equation is:
[tex]\[ 2 \, AgNO _3( aq ) + K _2 CrO _4( aq ) \rightarrow Ag _2 CrO _4(s) + 2 \, KNO _3( aq ) \][/tex]

In the previous step, you determined 0.040 mol of [tex]$K _2 CrO _4$[/tex] react.

The molar mass of [tex]$KNO _3$[/tex] is [tex]$101.11 \, g / mol$[/tex].

How many grams of [tex][tex]$KNO _3$[/tex][/tex] can form during the reaction?

[tex]\[ \text{Mass (g) } KNO _3 = \][/tex]



Answer :

To determine how many grams of KNO₃ can form, we proceed by following the stoichiometric relationships given in the balanced chemical equation:

[tex]\[ 2 \, \text{AgNO}_3 (aq) + \text{K}_2\text{CrO}_4 (aq) \rightarrow \text{Ag}_2\text{CrO}_4 (s) + 2 \, \text{KNO}_3 (aq) \][/tex]

1. Identify the molar ratios from the balanced equation:

From the balanced equation, we see that 1 mole of K₂CrO₄ reacts to produce 2 moles of KNO₃.

2. Determine the moles of K₂CrO₄ provided:

From the problem, we know that 0.040 mol of K₂CrO₄ react in the solution.

3. Calculate the moles of KNO₃ produced:

Using the stoichiometric ratio (1 mole of K₂CrO₄ produces 2 moles of KNO₃), we find:
[tex]\[ \text{Moles of KNO₃} = 2 \times 0.040 \, \text{mol} = 0.080 \, \text{mol} \][/tex]

4. Calculate the mass of KNO₃ formed:

The molar mass of KNO₃ is given as 101.11 g/mol. To find the mass, we use the formula:
[tex]\[ \text{Mass of KNO₃} = \text{moles of KNO₃} \times \text{molar mass of KNO₃} \][/tex]
Substituting the known values, we get:
[tex]\[ \text{Mass of KNO₃} = 0.080 \, \text{mol} \times 101.11 \, \text{g/mol} = 8.0888 \, \text{g} \][/tex]

Therefore, the mass of KNO₃ that can form during the reaction is 8.0888 grams.