How many kilograms of chromium can be obtained by the reaction of 236 kg of chromite with an excess of coke? Round of the answer to the nearest integer:

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Solution

$\begin{matrix} F e C r_{2} O_{4} (s) 1 m o l 224 g \end{matrix} + 4 C (s) \to F e (s) + \begin{matrix} 2 C r (s) 2 m o l 104 g \end{matrix} + \begin{matrix} 4 C O (g) 4 m o l 112 g \end{matrix}$
224 kg of chromite gives = 104 kg of Cr
Thus, 236 kg of chromite gives = $\frac{104 \times 236}{224} k g o f C r$
= 109.57 kg Cr

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How many kilograms of chromium can be obtained by the reaction of 236 kg of chromite with an excess of coke? Round of the answer to the nearest integer:

FeCr2O4(s)1mol224g+4C(s)→Fe(s)+2Cr(s)2mol104g+4CO(g)4mol112g224 kg of chromite gives = 104 kg of CrThus, 236 kg of chromite gives = 104×236224kgofCr = 109.57 kg Cr