Question

how much calcium carbonate can be prepared by 280 grams of calcium oxide?​

Answer 1

Answer:

Gram molecular mass of Cao is 56 gm. Amount of CaCO3 is 280 gm. Now, from 1 mole CaCO3, we can get 1 mole Cao. So 280/56=5 mole of CaCO3 can be prepared

Explanation:

Answer 2

Answer:

500 grams of calcium carbonate can be prepared by 280 grams of calcium oxide.

Explanation:

The reaction will be as follows,

CaO+CO₂ ⇒ CaCO₃                       (1)

Where,

CaO=Calcium oxide

CO₂=carbon dioxide

CaCO₃=calcium carbonate

The molar mass of calcium oxide is 56 grams and from equation (1) we can see that 56 grams of CaO are producing 100 grams of CaCO₃. So, 1 gram of  CaO will produce $\frac{100}{56}$  a gram of CaCO₃.So, the amount of calcium carbonate prepared by 280 grams of calcium oxide is calculated as,

$\frac{100}{56}\times 280=500$

Hence, 500 grams of calcium carbonate can be prepared by 280 grams of calcium oxide.

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