Question

chapter 11: skill - builder
due sep 14 by 11:59pm points 10
submitting an external tool
fill in the missing words
if 6.00 moles of $co_2$ have reacted with 6.00 moles of water in the photosynthesis reaction: $6co_2 + 6h_2o
ightarrow c_6h_{12}o_6 + 6o_2$, the volume of produced oxygen gas is

l assuming the standard temperature and pressure.

Explanation:

Step1: Determine the limiting reactant

Both $CO_2$ and $H_2O$ have 6.00 moles. From the balanced - equation $6CO_2 + 6H_2O
ightarrow C_6H_{12}O_6+6O_2$, the mole ratio of $CO_2$ to $H_2O$ is 1:1, so neither is in excess.

Step2: Calculate moles of $O_2$ produced

From the balanced equation, the mole ratio of $H_2O$ (or $CO_2$) to $O_2$ is 6:6 = 1:1. So, if 6.00 moles of $H_2O$ react, 6.00 moles of $O_2$ are produced.

Step3: Use the molar - volume relationship at STP

At standard temperature and pressure (STP), 1 mole of any ideal gas occupies 22.4 L. So, for 6.00 moles of $O_2$, the volume $V=n\times V_m$, where $n = 6.00$ moles and $V_m=22.4$ L/mol. Then $V = 6.00\times22.4$ L.
$V=134.4$ L

Answer:

134.4