1.

1.0 g of Mg is burnt with 0.28 g of O_(2)in a closed vessel. Which reactant is left in excess and how much ?

Answer»

Mg, 5.8 g
Mg, 0.58 g
`O_(2)`, 0.24g
`O_(2)`, 2.4g

Solution :`2Mg+O_(2) to 2MgO`
48G of Mg requires 32 g of `O_(2)`
? requires 0.28 g of `O_(2)`
Mass of magnesium REQUIRED `=(48xx0.28)/(32)=0.42g`
But 1 g of Mg is available. Thus, Mg is the excess REAGENT.
Excess of Mg LEFT BEHIND `=1-0.42=0.58g`


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