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inn [45]
3 years ago
13

What mass of H2SO4 is contained in 60.00 mL of a 5.85 M solution of sulfuric acid

Chemistry
1 answer:
strojnjashka [21]3 years ago
5 0

First, we have to calculate the number of moles of H2SO4 in the solution:

V=60 mL = 0.06 L

c=5.85 mol/L

n=V×c=0.06×5.85=0.351 mol

Then we need to find the molar mass of H2SO4:

2×Ar(H) + Ar(S) + 4×Ar(O) =

=2 + 32 + 64 = 98 g/mol

Finally, we need to find the mass of H2SO4:

m=0.351 × 98 = 34.398 g

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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.

4 0
4 years ago
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