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3 years ago

7

In the reaction 2AgI + Na2S → Ag2S +2NaI, calculate the number of moles of AgI needed to react with 85.0 g

1 answer:

cluponka [151]3 years ago

8 0

Answer:

8.5gNaI × 1molNaI/78gNaI × 234.7g AgI / 1 MOL AgI × 2Mol Ag I / 1mol Na2S=51.2 g of AgI reacts with Na2 S

Explanation:

conversion factor is used to make sure that all the units cancels with the other units and we only remain with the mass of Ag I which is 51.2g but since the number of moles can be calculated by dividing with the relative molecular mass therefore 51.2g/234.7gmol=0.22mol

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Explanation:

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<em>("volume must be in dm^3 , this is the reason radius is changed into dm"</em>

<em>"this is done because 1 dm^3 = 1 liter and concentration is always measured in liters")</em>

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