0.64 grams of H₂O
Given:
The combustion reaction of C₄H₆ consumes 2.1 grams of oxygen gas.
Question:
How many grams of H₂O is produced?
The Process:
Step-1
Let us convert mass to mol of 2.1 g of O₂.
[tex]\boxed{ \ n = \frac{mass}{Mr} \ } \rightarrow \boxed{ \ n = \frac{2.1}{32} = 0.0656 \ moles \ }[/tex]
Therefore, 0.0656 moles of O₂ are consumed.
Step-2
Balanced reaction:
[tex]\boxed{ \ 2C_4H_6_{(g)} + 11O_2_{(g)} \rightarrow 8CO_2_{(g)} + 6H_2O_{(g)} \ }[/tex]
Due to excess C₄H₆, then O₂ become the limiting reagent. The number of moles of O₂ determines the number of moles of H₂O as a result.
According to the chemical equation above, the proportion between O₂ and H₂O is 11 to 6. Therefore, we can count the number of moles of H₂O.
[tex]\boxed{ \ \frac{n(H_2O)}{n(O_2)} = \frac{6}{11} \ }[/tex] [tex]\ \rightarrow \boxed{ \ n(H_2O) = \frac{6}{11} \times n(O_2) \ }[/tex]
[tex]\boxed{ \ n(H_2O) = \frac{6}{11} \times 0.0656 \ }[/tex]
n (H₂O) = 0.0358 (round off to four decimal places)
Hence we get 0.0358 moles of H₂O.
Step-3
Let us convert mol to mass of 0.12 moles of H₂O.
[tex]\boxed{ \ n = \frac{mass}{Mr} \ } \rightarrow \boxed{ \ mass = n \times Mr \ }[/tex]
[tex]\boxed{ \ mass = 0.0358 \ moles \times 18 \ \frac{g}{mol} \ }[/tex]
[tex]\boxed{ \ mass = 0.64 \ grams \ }[/tex] (round off to two decimal place)
Thus, we can produce 0.64 grams of H₂O.
