Question

by using 8 grams for each of Ca and Br, how many grams of CaBr2 is possible to be produce?

B

Also determine the limiting and excess reagents.​

Answer 1

Answer:atomic mass of bromine is more .. so Mole of Br is less..

hence it is limiting agent.

Ca will be excess reagent

moles of Br2 = 8/160 = 1/20

hence 1/20 Moles of CaBr2 is formed

mass = 1/20 x 200= 10g

Explanation:

Answer 2

Explanation:

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