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We wish to determine how many grams of [tex]$KNO_3$[/tex] can form when 100 mL of [tex]$0.40 \, \text{M} \, K_2CrO_4$[/tex] solution is added to excess [tex][tex]$AgNO_3$[/tex][/tex].

The balanced chemical equation is:
[tex]
2 \, AgNO_3(aq) + K_2CrO_4(aq) \rightarrow Ag_2CrO_4(s) + 2 \, KNO_3(aq)
[/tex]

How many moles of [tex]$K_2CrO_4$[/tex] are present in 100 mL of [tex]$0.40 \, \text{M} \, K_2CrO_4$[/tex]?

[tex]\text{Enter answer in moles of } K_2CrO_4.[/tex]

Sagot :

GinnyAnswer
Certainly! Let's start by calculating the number of moles of [tex]\( K_2CrO_4 \)[/tex] present in 100 mL of a 0.40 M [tex]\( K_2CrO_4 \)[/tex] solution.

1. Understanding the given values:
- Volume of [tex]\( K_2CrO_4 \)[/tex] solution: [tex]\( 100 \)[/tex] mL
- Concentration of [tex]\( K_2CrO_4 \)[/tex] solution: [tex]\( 0.40 \)[/tex] M

2. Convert the volume from milliliters to liters:
[tex]\( 100 \)[/tex] mL = [tex]\( 0.100 \)[/tex] L

3. Apply the formula to find moles from volume and concentration:
The formula to calculate the number of moles is:
[tex]\[ \text{moles} = \text{volume (L)} \times \text{concentration (M)} \][/tex]
Substituting the given values:
[tex]\[ \text{moles of } K_2CrO_4 = 0.100 \, \text{L} \times 0.40 \, \text{M} \][/tex]

4. Calculate the number of moles:
[tex]\[ \text{moles of } K_2CrO_4 = 0.100 \times 0.40 = 0.040 \, \text{moles} \][/tex]

Therefore, the number of moles of [tex]\( K_2CrO_4 \)[/tex] present in 100 mL of a 0.40 M [tex]\( K_2CrO_4 \)[/tex] solution is [tex]\( 0.040 \)[/tex] moles.
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