consider the balanced chemical equation for the combustion of methane (ch₄). ch₄(g) + 2o₂(g)→co₂(g) +…

consider the balanced chemical equation for the combustion of methane (ch₄). ch₄(g) + 2o₂(g)→co₂(g) + 2h₂o(g) given that the molar mass of co₂ is 44.01 g/mol, how many liters of oxygen is required at stp to produce 88.0 g of co₂ from this reaction? 44.8 l 45.00 l 89.55 l 89.6 l

consider the balanced chemical equation for the combustion of methane (ch₄). ch₄(g) + 2o₂(g)→co₂(g) + 2h₂o(g) given that the molar mass of co₂ is 44.01 g/mol, how many liters of oxygen is required at stp to produce 88.0 g of co₂ from this reaction? 44.8 l 45.00 l 89.55 l 89.6 l

Answer

Explanation:

Step1: Calculate moles of CO₂

First, use the formula $n=\frac{m}{M}$, where $m$ is mass and $M$ is molar - mass. Given $m = 88.0\ g$ and $M = 44.01\ g/mol$ of $CO_2$. $n_{CO_2}=\frac{88.0\ g}{44.01\ g/mol}\approx2\ mol$

Step2: Determine moles of O₂ from stoichiometry

From the balanced chemical equation $CH_4(g)+2O_2(g)\rightarrow CO_2(g) + 2H_2O(g)$, the mole - ratio of $O_2$ to $CO_2$ is $2:1$. So, if $n_{CO_2}=2\ mol$, then $n_{O_2}=2\times n_{CO_2}=4\ mol$.

Step3: Calculate volume of O₂ at STP

At STP (Standard Temperature and Pressure, $T = 273\ K$ and $P = 1\ atm$), the molar volume of a gas is $V_m = 22.4\ L/mol$. Use the formula $V=n\times V_m$. $V_{O_2}=n_{O_2}\times V_m=4\ mol\times22.4\ L/mol = 89.6\ L$

Answer:

D. 89.6 L