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matrenka [14]
2 years ago
15

how many gallons of water did a truck fill up if it's weight before the fill up was 3547 lbs and afterwards was 4924 lbs?

Chemistry
1 answer:
musickatia [10]2 years ago
3 0

Answer: depending on the method, 138-139 (imperial) gallons, or 626 L

Explanation:

The wt of water is 4924-3547 lb = 1377 lb = 1377/2.2 kg = 626 kg = 626 L

1gallon = 4.5 L (it does where I come from and who still measures things in imperial anyway?) so I guess  that is 139 gallons

or alternatively, I recall from distant childhood, “a pint of water weighs a pound and a quarter” which means 1 gallon = 10 lb, so 1377 lb = ~138 gallons

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Mariana [72]

Answer:

H₂SO₄

Explanation:

Given data:

Number of moles of H₂SO₄ = 15 mol

Number of moles of Fe = 13 mol

Which reactant is limiting reactant = ?

Solution:

Chemical equation:

3H₂SO₄  + 2Fe      →          Fe₂(SO₄)₃  + 3H₂

now we will compare the moles reactant with product.

               H₂SO₄         :          Fe₂(SO₄)₃  

                  3               :              1

                 15               :              1/3×15 = 5

                H₂SO₄         :            H₂

                  3               :              3

                 15               :              15

                Fe               :          Fe₂(SO₄)₃  

                  2               :              1

                 13               :              1/2×13 = 6.5

                Fe               :                H₂

                  2               :                 3

                 13               :              3/2×13 = 19.5

Number of moles of product formed by  H₂SO₄ are less thus it will act as limiting reactant.

8 0
3 years ago
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Veseljchak [2.6K]

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4 0
3 years ago
For the following reaction, 4.21 grams of hydrogen gas are allowed to react with 10.6 grams of ethylene (C2H4) . hydrogen(g) + e
sergiy2304 [10]

Answer:a)  11.34 g of ethane (C_2H_6) can be formed

b) C_2H_4 is the limiting reagent

c) 3.44 g of the excess reagent remains after the reaction is complete

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}    

1. \text{Moles of} H_2=\frac{4.21}{2}=2.10moles

2. \text{Moles of} C_2H_4=\frac{10.6}{28}=0.378moles

H_2(g)+C_2H_4(g)\rightarrow C_2H_6(g)

According to stoichiometry :

1 mole of C_2H_4 require 1 mole of H_2

Thus 0.378 moles of C_2H_4 will require=\frac{1}{1}\times 0.378=0.378moles  of H_2

Thus C_2H_4 is the limiting reagent as it limits the formation of product and H_2 is the excess reagent.

moles of H_2 left = (2.10-0.378) = 1.72 moles

mass of H_2 left=moles\times {\text {Molar mass}}=1.72moles\times 2g/mol=3.44g

According to stoichiometry :

As 1 mole of C_2H_4 give = 1 mole of C_2H_6

Thus 0.378 moles of C_2H_4 give =\frac{1}{1}\times 0.378=0.378moles  of C_2H_6

Mass of C_2H_6=moles\times {\text {Molar mass}}=0.378moles\times 30g/mol=11.34g

Thus 11.34 g of ethane is formed.

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