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3 years ago

How many lbs. of O2 are required to fully convert 1-lb of glucose (C6H12O6) to carbon dioxide and water

Chemistry

1 answer:

Serjik [45]3 years ago

3 0

Answer:

The correct answer is : 1.07 lbs

Explanation:

solution:

molar mass glucose (C6H12O6) = 180 g/mol

molar mass of oxygen molecule (O2) = 32 g/mol (as we know molar mass of O = 16 g/mol)

the balanced reaction of conversion of water and oxygen to glucose is:

$C_{6}H_{12}O_{6} + 6O_{2} \rightarrow 6CO_{2} + 6H_{2}O ...(A)$

1 mol of C6H12O6 = 6 mol of O2 (from reaction A)

so, 180 g C6H12O6 = 192 g O2

that is, 0.396 lb of C6H12O6 = 0.423 lb of O2 ( 1 g = 0.00220462 lb )

so,

1 lb C6H12O6 = $\frac{1(lb) \times 0.4233 (lb)}{0.3968 (lb)}$ = 1.07 lb O2

therefore, the correct answer is : 1.07 lbs

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What would cause the percent yield of a reaction to be less than 100%?

Roman55 [17]

<u>Answer:</u>

All of the limiting reagent gets used up causes the percent yield of a reaction to be less than 100%

<u>Explanation:</u>

It is rare for a chemical reaction to occur with the right proportion of reactants that reacts together to form resultant products.During this process, one of the reactants gets used up faster resulting in the lower percent yield. Such product that gets used up faster is called limiting reagent. This limiting reagent is responsible for lowering the percentage yield in the chemical reaction.

The limiting reactant can be found easily by calculating the yield of each reactant assuming they are consumed completely. The reactant which has least yield is the limiting reactant of the reaction.

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3 years ago

Explain what might happen to tundra animals, such as polar bears, as earth's climate warms. pls explain

SCORPION-xisa [38]

Answer:

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2 years ago

6.0 mol NaOH reacts with

lina2011 [118]

Taking into account the reaction stoichiometry, 2 moles of Na₃PO₄ can be produced when 6.0 mol NaOH reacts with 9.0 mol H₃PO₄.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

3 NaOH + H₃PO₄ → 3 H₂O + Na₃PO₄

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

NaOH: 3 moles
H₃PO₄: 1 mole
H₂O: 3 moles
Na₃PO₄: 1 mole

<h3>Limiting reagent</h3>

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

<h3>Limiting reagent in this case</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 1 mole of H₃PO₄ reacts with 3 moles of NaOH, 9 moles of H₃PO₄ reacts with how many moles of NaOH?

$moles of NaOH=\frac{9 moles of H_{3} PO_{4} x3 moles of NaOH}{1 mole of H_{3} PO_{4}}$

moles of NaOH= 27 moles

But 27 moles of NaOH are not available, 6 moles are available. Since you have less moles than you need to react with 9 moles of H₃PO₄, NaOH will be the limiting reagent.

<h3>Moles of Na₃PO₄ formed</h3>

Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 3 moles of NaOH form 1 mole of Na₃PO₄, 6 moles of NaOH form how many moles of Na₃PO₄?

$moles of Na_{3}P O_{4} =\frac{6 moles of NaOHx1 mole of Na_{3}P O_{4} }{3 moles of NaOH}$