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2 years ago

What mass of carbon monoxide must be burned to produce 175 kj of heat under standard state conditions?

Chemistry

2 answers:

Soloha48 [4]2 years ago

8 0

When 2.50 g is burned then in oxygen then 1.25kj of heat is produced.

Mrrafil [7]2 years ago

7 0

Answer:

17.3 g

Explanation:

Let's consider the combustion of CO.

CO(g) + 0.5 O₂(g) → CO₂(g)

We can find the standard enthalpy of the reaction using the following expression.

ΔH°rxn = 1 mol × ΔH°f(CO₂(g)) - 1 mol × ΔH°f(CO(g)) - 0.5 mol × ΔH°f(O₂(g))

ΔH°rxn = 1 mol × (-393.5 kJ/mol) - 1 mol × (-110.5 kJ/mol) - 0.5 mol × 0

ΔH°rxn = -283.0 kJ

283.0 kJ are released per mole of CO. The moles of CO required to release 175 kJ are:

(-175 kJ) × (1 mol CO/ -283.0 kJ) = 0.618 mol CO

The molar mass of CO is 28.01 g/mol. The mass of CO is:

0.618 mol × (28.01 g/mol) = 17.3 g

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If the actual yield of a reaction is 37.6 g while the theoretical yield is 112.8 g what is the percent yield

Zigmanuir [339]

<h2>Hello!</h2>

The answer is:

The percent yield of the reaction is 32.45%

To calculate the percent yield, we have to consider the theoretical yield and the actual yield. The theoretical yield as its name says is the yield expected, however, many times the difference between the theoretical yield and the actual yield is notorious.

We are given that:

$ActualYield=37.6g\\TheoreticalYield=112.8g$

Now, to calculate the percent yield, we need to divide the actual yield by the theoretical and multiply it by 100.

So, calculating we have:

$PercentYield=\frac{ActualYield}{TheoreticalYield}*100\\\\PercentYield=\frac{37.6g}{112.8g}*100=0.3245*100=32.45(percent)$

Hence, we have that the percent yield of the reaction is 32.45%.

Have a nice day!

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