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Norma-Jean [14]
3 years ago
5

A new grill has a mass of 30.0 kg. you put 1.5 kg of charcoal in the grill. you burn all the charcoal and the grill has a mass o

f 30.0 kg. what is the mass in kg of the gases given off? (assume that the charcoal is pure carbon solid and that it burns completely in oxygen).
Chemistry
1 answer:
atroni [7]3 years ago
7 0

The reaction for burning of charcoal or complete combustion is as follows:

C(s)+O_{2}(g)\rightarrow CO_{2}(g)

From the above balanced reaction, 1 mole of carbon releases 1 mole of CO_{2} gas.

Converting mass of charcoal into moles as follows:

n=\frac{m}{M}

Molar mass of pure carbon is 12 g/mol thus,

n=\frac{1.5\times 10^{3} g}{12 g/mol}=125mol

The same moles of CO_{2} is released. Converting these moles into mass as follows:

m=n×M

Molar mass of CO_{2} is 44 g/mol thus,

m=125mol\times 44 g/mol=5.5\times 10^{3}g

Converting mass into kg,

1g=10^{-3}kg

Thus, total mass of gas released is 5.5 kg.

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What is the volume of 45.0 grams of nitrogen monoxide , NO
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Answer:

V=37.05\ L

Explanation:

Given that:

Mass of NO, m = 45.0 g

Molar mass of NO, M = 30.01 g/mol

Temperature = 20.0 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (20.0 + 273.15) K = 293.15 K

V = ?

Pressure = 740 mm Hg

Considering,  

n=\frac{m}{M}

Using ideal gas equation as:

PV=\frac{m}{M}RT

where,  

P is the pressure

V is the volume

m is the mass of the gas

M is the molar mass of the gas

T is the temperature  

R is Gas constant having value = 62.36367 L. mmHg/K. mol

Applying the values in the above equation as:-

740\times V=\frac{45.0}{30.01}\times 62.36367\times 293.15

740V=\frac{822685.94372}{30.01}

V=37.05\ L

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3 years ago
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Use a sheet of paper to answer the following question. Take a picture of your answers and attach to this assignment. Predict the
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How many grams of Na2SO4 will be formed from 200.0g of NaOH
nikklg [1K]

The question is incomplete, the complete question is;

Using the following equation 2 NaOH(aq) + H2SO4(aq) → 2 H2O(aq) + Na2SO4(aq) how many grams of sodium sulfate will be formed if you start with 200 grams of sodium hydroxide and you have an excess of sulfuric acid

Answer:

355.1 g

Explanation:

The equation of the reaction is;

2 NaOH(aq) + H2SO4(aq) → 2 H2O(aq) + Na2SO4(aq)

We have been told that H2SO4 is in excess so NaOH is the limiting reactant. Therefore;

Number of moles in 200g of NaOH = 200g/40g/mol = 5 moles

So;

2 moles of NaOH yields 1 mole of Na2SO4

5 moles of NaOH will yield 5 * 1/2 = 2.5 moles of Na2SO4

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