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Consider the following reaction
$\to_{{Br}_{2}}^{redphosphorous} R (majorproduct)$
On estimation of bromine in 1.00 g of R using Carius method, the amount of AgBr formed (in g) is ________.
[Given: Atomic mass of H = 1, C = 12, O = 16, P = 31, Br = 80, Ag = 108]

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Detailed Solution

$2 P + 3 {Br}_{2} \to 2 {PBr}_{3}$
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Number of moles in 1 gm of (R) = $\frac{1}{250}$
Number of moles of AgBr formed from (R) = $\frac{2}{250}$
Mass of AgBr formed = $= \frac{2 \times 188}{250} = 1.50 gm$

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