Question

learn by doing
the fermentation of sugar (glucose) to alcohol (ethanol) occurs according to the following reaction.
·c₆h₁₂o₆(s) → 2 c₂h₅oh(l) + 2 co₂(g)
if 9.22 grams of glucose reacts, how many liters of carbon dioxide will be produced at stp?
● 2.29 l
● 0.575 l
● 4.52 l
● 18.5 l
did i get this
potassium is an extremely reactive element. it reacts violently with water. the equation for the
reaction is seen below.
2 k(s) + 2 h₂o(l) → 2 koh (aq) + h₂(g)
if 2.18 g of potassium reacts with an excess amount of water, what volume of h₂ will be produced
at stp? l

Explanation:

Step1: Calculate moles of glucose

The molar mass of \(C_6H_{12}O_6\) is \(180.16\ g/mol\).
\(n_{C_6H_{12}O_6}=\frac{m}{M}=\frac{9.22\ g}{180.16\ g/mol}\approx0.0512\ mol\)

Step2: Find moles of \(CO_2\)

From the reaction \(C_6H_{12}O_6(s)\to2C_2H_5OH(l)+2CO_2(g)\), the mole ratio of \(C_6H_{12}O_6\) to \(CO_2\) is \(1:2\).
\(n_{CO_2}=2\times n_{C_6H_{12}O_6}=2\times0.0512\ mol = 0.1024\ mol\)

Step3: Calculate volume of \(CO_2\) at STP

At STP (\(T = 273\ K\), \(P=1\ atm\)), \(V = n\times V_m\), where \(V_m = 22.4\ L/mol\)
\(V_{CO_2}=n_{CO_2}\times22.4\ L/mol=0.1024\ mol\times22.4\ L/mol\approx2.29\ L\)

Answer:

\(2.29\ L\)