65 m^3 are fed / h of benzene to a reactor, it is requested: to. What is the mass flow fed in kg / h? b. And the molar flow in m
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The answer for the following problem is mentioned below.
<u><em>Therefore volume occupied by methane gas is 184.78 × 10^-3 liters </em></u>
Explanation:
Given:
mass of methane() = 272 grams
pressure (P) = 250 k Pa =250×10^3 Pa
temperature(t) = 54°C =54 + 273 = 327 K
Also given:
R = 8.31JK-1 mol-1 ,
Molar mass of methane() = 16.0 grams
We know;
According to the ideal gas equation,
<u><em>P × V = n × R × T</em></u>
here,
n = m÷M
n =272 ÷ 16
<u><em>n = 17 moles</em></u>
Therefore,
250×10^3 × V = 17 × 8.31 × 327
V = ( 17 × 8.31 × 327 ) ÷ ( 250×10^3 )
V = 184.78 × 10^-3 liters
<u><em>Therefore volume occupied by methane gas is 184.78 × 10^-3 liters </em></u>
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Answer:
<u>STEP I</u>
This is the balanced equation for the given reaction:-
<u>STEP II</u>
The compounds marked with (aq) are soluble ionic compounds. They must be
broken into their respective ions.
see, in the equation KOH, H2SO4, and K2SO4 are marked with (aq).
On breaking them into their respective ions :-
- 2KOH -> 2K+ + 2OH-
- H2SO4 -> 2H+ + (SO4)2-
- K2SO4 -> 2K+ + (SO4)2-
<u>STEP III</u>
Rewriting these in the form of equation
<u>STEP </u><u>IV</u>
Canceling spectator ions, the ions that appear the same on either side of the equation
<em>(note: in the above step the ions in bold have gotten canceled.)</em>
This is the net ionic equation.
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- KOH has been taken as aqueous because the question informs us that we have a solution of KOH. by solution it means that KOH has been dissolved in water before use.
[Alkali metal hydroxides are the only halides soluble in water ]