Calculate the mass of silver bromide produced from 22.5 g of silver nitrate.
1 answer:
Equation is as follow,
<span> 2 AgNO</span>₃<span> + MgBr</span>₂<span> </span>→ <span>2 AgBr + Mg(NO</span>₃<span>)</span>₂
According to eq.
339.74 g (2 moles) AgNO₃ produces = 375.54 g (2 moles) of AgBr
So,
22.5 g AgNO₃ will produce = X g of AgBr
Solving for X,
X = (22.5 g × 375.54 g) ÷ 339.74 g
X = 24.87 g of AgBr
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