Question

Question 22

When 85 g of CH4 are mixed with 320 g of O2 the limiting reactant is
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

Answer 1

O2, oxygen.

Step-by-step explanation:

Hello.

In this case, for the undergoing chemical reaction, we need to compute the moles of CO2 yielded by 85 g of CH4 (molar mass = 16 g/mol) and by 320 g of O2 (molar mass 32 g/mol) via the following mole-mass relationships:

`n_(CO_2)^(by\ CH_4)=85gCH_4*(1molCH_4)/(16gCH_4) *(1molCO_2)/(1molCH_4) =5.3molCO_2\\\\n_(CO_2)^(by\ O_2)=320gO_2*(1molO_2)/(32gO_2) *(1molCO_2)/(2molO_2) =5molCO_2`

Considering the 1:2:1 among CH4, O2 and CO2. Therefore, since 320 g of O2 yield the smallest amount of CO2 we infer that the limiting reactant is O2.

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