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3 years ago

If 155 grams of potassium (K) reacts with 122 grams of potassium nitrate (KNO3), what is the limiting reagent?

1 answer:

5 0

K:

m=155g
M=39g/mol

n = 155g / 39g/mol ≈ 3,97mol

KNO₃:

m=122g
M=101g/mol

n = 122g/101g/mol = 1,21mol

2K + 10KNO₃ ⇒ 6K₂O + N₂
2mol : 10mol
3,97mol : 1,21mol
limiting reagent

KNO₃ is limiting reagent

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Answer:

Explanation:

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Answer:

$\boxed{\text{ B. Increase the temperature and decrease the pressure.}}$

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Another way to write the equilibrium would be

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Let's consider each of the stresses in turn.

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3 years ago

Read 2 more answers

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Mama L [17]

Answer:

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Explanation:

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3 years ago

A mixture of methanol and methyl acetate contains 15.0 wt% methanol.

Usimov [2.4K]

The mixture flow rate in lbm/h = 117.65 lbm/h

<h3>Further explanation</h3>

Given

15.0 wt% methanol

The flow rate of the methyl acetate :100 lbm/h

Required

the mixture flow rate in lbm/h

Solution

mass of methanol(CH₃OH, Mw= 32 kg/kmol) in mixture :

$\tt 15\%\times 200~kg=30~kg\\\\mol=\dfrac{mass}{MW}=\dfrac{30~kg}{32~kg/kmol}=0.9375~kmol$

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$\tt 85\%\times 200=170~kg\\\\mol=\dfrac{170}{74}=2.297~kmol$

Flow rate of the methyl acetate in the mixture is to be 100 lbm/h.

1 kg mixture = 0.85 .methyl acetate

So flow rate for mixture :

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3 years ago