# How many ml of .150 M Na2CrO4 will be required to oxidise 40 ml of .05M Na2S2O3

## To determine the volume of 0.150 M Na2CrO4 required to oxidize 40 mL of 0.05 M Na2S2O3, we need to use the stoichiometry of the balanced chemical equation for the reaction between Na2CrO4 and Na2S2O3.

The balanced equation for the reaction is:

2 Na2S2O3 + Na2CrO4 → Na2S4O6 + Na2CrO7

From the balanced equation, we can see that the stoichiometric ratio between Na2S2O3 and Na2CrO4 is 2:1. This means that for every 2 moles of Na2S2O3, we need 1 mole of Na2CrO4.

1 mole of Na2S2O3 is equal to its molarity (0.05 M) multiplied by its volume in liters:

1 mole = 0.05 M x 0.040 L
= 0.002 moles

Since the stoichiometry ratio is 2:1, we need half as many moles of Na2CrO4:

0.002 moles Na2CrO4

To calculate the volume of 0.150 M Na2CrO4, we need to convert moles to volume using the molarity:

Volume (L) = Moles / Molarity

Volume (L) = 0.002 moles / 0.150 M
= 0.0133 L

Finally, to convert the volume to milliliters, we multiply by 1000:

Volume (mL) = 0.0133 L x 1000
= 13.3 mL

Therefore, 13.3 mL of 0.150 M Na2CrO4 will be required to oxidize 40 mL of 0.05 M Na2S2O3.

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## To find out how many mL of 0.150 M Na2CrO4 are required to oxidize 40 mL of 0.05 M Na2S2O3, we need to use the concept of stoichiometry and the balanced chemical equation.

The balanced chemical equation for the reaction between Na2CrO4 (sodium chromate) and Na2S2O3 (sodium thiosulfate) is:

2 Na2CrO4 + 3 Na2S2O3 -> Na4Cr2O7 + 3 Na2S4O6

From the balanced equation, we can see that the ratio of Na2CrO4 to Na2S2O3 is 2:3, meaning that 2 moles of Na2CrO4 react with 3 moles of Na2S2O3.

First, let's calculate the number of moles of Na2S2O3 present in the 40 mL of 0.05 M solution:

Number of moles of Na2S2O3 = Volume (in L) x Concentration (in M)
= 0.040 L x 0.05 M
= 0.002 moles

Since the ratio of Na2CrO4 to Na2S2O3 is 2:3, we can find the number of moles of Na2CrO4 required:

Number of moles of Na2CrO4 = (3/2) x Number of moles of Na2S2O3
= (3/2) x 0.002 moles
= 0.003 moles

Now, we need to find the volume of the 0.150 M Na2CrO4 solution required to have 0.003 moles:

Volume (in L) = Number of moles / Concentration (in M)
= 0.003 moles / 0.150 M
= 0.02 L

Finally, we convert the volume from liters to milliliters:

Volume (in mL) = Volume (in L) x 1000
= 0.02 L x 1000
= 20 mL

Therefore, 20 mL of a 0.150 M Na2CrO4 solution will be required to oxidize 40 mL of a 0.05 M Na2S2O3 solution.

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