Chemistry

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 ]

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

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

Chemical Equation:

$\underset{170 g}{\text{AgNO}_3} + \text{HCl} \longrightarrow \underset{143.5 g}{\text{AgCl}} + \text{HNO}_{3}$

170 g of AgNO₃ gives 143.5 g of AgCl

∴ 0.34 g of AgNO₃ will give $\dfrac{143.5}{170}$ x 0.34 = 0.287 g of AgCl

Hence, 0.287 g of silver chloride will be obtained

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