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3 years ago

PLEASE HELP

Chemistry

1 answer:

swat323 years ago

5 0

Answer:

1461.7 g of AgI

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

CaI₂ + 2AgNO₃ —> 2AgI + Ca(NO₃)₂

From the balanced equation above,

1 mole of CaI₂ reacted to produce 2 moles of AgI.

Next, we shall determine the number of mole AgI produced by the reaction of 3.11 moles of CaI₂. This can be obtained as follow:

From the balanced equation above,

1 mole of CaI₂ reacted to produce 2 moles of AgI.

Therefore, 3.11 moles of CaI₂ will react to produce = 3.11 × 2 = 6.22 moles of AgI

Finally, we shall determine the mass of 6.22 moles of AgI. This can be obtained as follow:

Mole of AgI = 6.22 moles

Molar mass of AgI = 108 + 127

= 235 g/mol

Mass of AgI =?

Mass = mole × molar mass

Mass of AgI = 6.22 × 235

Mass of AgI = 1461.7 g

Therefore, 1461.7 g of AgI were obtained from the reaction.

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3000 lb of 13% solution is required .

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n = moles of water / ( moles of water + moles of glucose)
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Pv(solu) = n Pv(water)
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