Chemistry, asked by shreyansh2141, 1 year ago

The weight of slaked lime required to decompose 8.0g NH4Cl is

Answers

Answered by ug57125

0

74 g of Slaked lime is needed for 107 g of NH4Cl

Mass of Slaked lime required for 8 g of NH4Cl =

74 x 8 / 107 = 5.53 g

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