if 0.683 grams of silver chloride is produced how much (mass) silver nitrate would need to be reacted

1 answer:

4 0

The (mass) silver nitrate would needed to be reacted is calculated as follow

by assuming that AgNo3 reacted with Bacl

that is 2AgNo3 + BaCl2 ---> 2Agcl + Ba(No3)2

find the moles AgNo3 = 0.683g/169.87 g/mol =4.02 x10^-3 moles

by use of mole ratio between Agcl and Agno3 which is 2:2 this implies that the moles of AgNO3 also 4.02 x10^-3

mass= moles x molar mass

= (4.02 x10^-3) x 143.37= 0.576 grams

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