Question

What mass of silver chloride will be obtained by adding an excess of hydrochloric acid to a solution of 0.34 g of silver nitrate. [Cl = 35.5, Ag = 108, N = 14, O = 16, H = 1 ]

August 2, 2024

Answer

Molecular Weight of Silver Nitrate (AgNO₃) = 108 + 14 + 48 = 170 g

Molecular Weight of Silver Chloride (AgCl) = 108 + 35.5 = 143.5 g

Balanced Chemical Equation:

AgNO₃ + HCl → AgCl + HNO₃

170 g 143.5 g

170 g of AgNO₃ gives 143.5 g of AgCl

∴ 0.34 g of AgNO₃ will give $143.5\over 170$ × 0.34

= 0.287 g of AgCl

Hence, 0.287 g of silver chloride will be

💡 Some Related Questions

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