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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 M K _2 CrO _4[/tex] solution is added to excess [tex]AgNO _3[/tex].

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

How many moles of [tex]K _2 CrO _4[/tex] are present in 100 mL of [tex]0.40 M K _2 CrO _4[/tex]?

Sagot :

GinnyAnsweridn
To determine how many moles of [tex]\( K_2CrO_4 \)[/tex] are present in 100 mL of 0.40 M [tex]\( K_2CrO_4 \)[/tex], let's follow these steps:

1. Understand the given data:
- Volume of [tex]\( K_2CrO_4 \)[/tex] solution = 100 mL
- Concentration of [tex]\( K_2CrO_4 \)[/tex] solution = 0.40 M (moles per liter)

2. Convert the volume from mL to L:
- Volume in liters = [tex]\(\frac{100\text{ mL}}{1000}\text{ L} = 0.100\text{ L}\)[/tex]

3. Use the definition of molarity (M) which is moles per liter (mol/L) to find the number of moles of [tex]\( K_2CrO_4 \)[/tex]:
- Moles of [tex]\( K_2CrO_4 \)[/tex] = Molarity [tex]\(\times\)[/tex] Volume in liters
- Moles of [tex]\( K_2CrO_4 \)[/tex] = 0.40 M [tex]\(\times\)[/tex] 0.100 L = 0.040 moles

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