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The answer is $67.6$ $g$.

First, start with the balanced chemical equation

$Bi_2O_(3(s)) + 3C_((s)) -> 2Bi_((s)) + 3CO_((g))$

We know the molar masses of $Bi$ ($209g/(mol)$) and $Bi_2O_3$ ($466g/(mol)$), and the between $Bi$ and $Bi_2O_3$, so we could say that

$60.7gBi* (1 mol e Bi)/(209g) * (1 mol e Bi_2O_3)/(2 mol es Bi) * (466g)/(1 mol e Bi_2O_3) = 67.6g Bi_2O_3$

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