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DanielleElmas [232]
3 years ago
8

How many grams of the excess reactant are left over according to the reaction below given that you start with 10.0 g of Al and 1

9.0 grams of O2?
Reaction: 2Al+ 3O2 → 2Al2O3
Chemistry
2 answers:
valentinak56 [21]3 years ago
7 0
<span>4 Al + 3 O2 → 2 Al2O3 

(10.0 g Al) / (26.98154 g Al/mol) = 0.37062 mol Al 
(19.0 g O2) / (31.99886 g O2/mol) = 0.59377 mol O2 

0.37062 mole of Al would react completely with 0.37062 x (3/4) = 0.277965 mole of O2, but there is more O2 present than that, so O2 is in excess. 

((0.59377 mol O2 initially) - (0.277965 mol O2 reacted)) x (31.99886 g O2/mol) = 
10.1 g O2 left over</span><span>
</span>
expeople1 [14]3 years ago
7 0

Answer : 10.1  G O2 .

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The volume of the 0.15 M LiOH solution required to react with 50 mL of 0.4 M HCOOH to the equivalence point is 133.3 mL

<h3>Balanced equation </h3>

HCOOH + LiOH —> HCOOLi + H₂O

From the balanced equation above,

The mole ratio of the acid, HCOOH (nA) = 1

The mole ratio of the base, LiOH (nB) = 1

<h3>How to determine the volume of LiOH </h3>
  • Molarity of acid, HCOOH (Ma) = 0.4 M
  • Volume of acid, HCOOH (Va) = 50 mL
  • Molarity of base, LiOH (Mb) = 0.15 M
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MaVa / MbVb = nA / nB

(0.4 × 50) / (0.15 × Vb) = 1

20 / (0.15 × Vb) = 1

Cross multiply

0.15 × Vb = 20

Divide both side by 0.15

Vb = 20 / 0.15

Vb = 133.3 mL

Thus, the volume of the LiOH solution needed is 133.3 mL

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A gray whale can travelan average of 120<br> km per day as it migrates.
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The average pea weighs between 0.1 and 0.36 grams.

If we take the lower value (0.1 g/pea), the number of peas in 454 g is:

454 g \times \frac{1pea}{0.1 g} = 4540pea

If we take the higher value (0.36 g/pea), the number of peas in 454 g is:

454 g \times \frac{1pea}{0.36 g} \approx  1261pea

​In a bag of peas that weighs 454 grams, there are between 1261 and 4540 peas.

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Answer:

b

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