Answer:
[tex]m_{CO_2}=33gCO_2[/tex]
Explanation:
Hello,
In this case the undergoing chemical reaction is:
[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]
Next we identify the limiting reactant with the given amounts of methane and oxygen by computing the available moles of methane and the moles of methane consumed by the 133 g of oxygen:
[tex]n_{CH_4}^{available}=12g*\frac{1mol}{16g} =0.75molCH_4\\n_{CH_4}^{consumed\ by\ O_2}=133gO_2*\frac{1molO_2}{32gO_2}*\frac{1molCH_4}{2molO_2}=2.08molCH_4[/tex]
In such a way, the limiting reactant is methane with 0.75 moles which produce the following mass of carbon dioxide:
[tex]m_{CO_2}=0.75molCH_4*\frac{1molCO_2}{1molCH_4} *\frac{44gCO_2}{1molCO_2} \\\\m_{CO_2}=33gCO_2[/tex]
Best regards.
Answer:
The correct answer is 33 g CO₂
Explanation:
The reaction between CH₄ and O₂ is a combustion reaction. The complete combustion reaction is described by the following chemical equation:
CH₄(g) + 2 O₂(g)→ CO₂(g) + 2 H₂O(g)
16 g 64 g 44 g 36 g
That means that 1 mol of CH₄ (equal to 16 g) reacts with 2 moles of O₂ (65 g) to give 1 mol of CO₂ (44 g) and 2 moles of H₂O (36 g).
In order to calculate the grams of CO₂ produced by the reaction, we have to figure out which is the limiting reactant. For this, we can choose one reactant and calculate how many grams we need of the other reactant by using the quantity of reactant we have. Let's choose CH₄. According to the chemical equation, 16 g of CH₄ reacts with 64 g of O₂. If we have 12 g, the quantity of O₂ we need is the following:
g O₂ requested = 12 g CH₄ x (64 g O₂)/(16 g CH₄) = 48 g
We need 48 g of O₂ and we have 133 g, so the O₂ is the excess reactant and CH₄ is the limiting reactant. With the limiting reactant, we calculate the amount of product produced. According to the chemical equation, with 16 g of CH₄ we obtain 44 g of CO₂. We have 12 g, so we multiply the grams of CH₄ we have by the factor 44 g CO₂/16 g CH₄:
grams of CO₂ produced = 12 g CH₄ x 44 g CO₂/16 g CH₄ = 33 g CO₂
