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

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Answer:

  1. 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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