$C_{2} H_{4} + 3 O_{2} \to 2 C O_{2} + 2 H_{2} O$ $C_2H_4 + 3O_2 -> 2CO_2 +2H_2O$ . If you start with 45 grams of ethylene ( $C_{2} H_{4}$ $C_2H_4$ ), how many grams of carbon dioxide will be produced?

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Answer 1 · 2017-03-29T18:20:02

141.416g

$No. of mol of ethylene = \frac{45}{2 \cdot 12 + 4 \cdot 1} \approx 1.607 m o l$ $"No. of mol of ethylene"=45/(2*12+4*1)~~1.607mol$

$No. of mol of C O_{2} produced = 2 \cdot 1.607 = 3.214 m o l$ $"No. of mol of "CO_2" produced"=2*1.607=3.214mol$

$Mass of C O_{2} = 3.214 \cdot (12 + 2 \cdot 16) = 141.416 g$ $"Mass of "CO_2"=3.214*(12+2*16)=141.416g$

C2H4+3O2→2CO2+2H2OC_2H_4 + 3O_2 -> 2CO_2 +2H_2O. If you start with 45 grams of ethylene (C2H4C_2H_4), how many grams of carbon dioxide will be produced?