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3 years ago

What is the theoretical yield of methanol (CH3OH) when 12.0 grams of H2 is mixed with 74.5 grams of CO? CO + 2H2 CH3OH

1 answer:

3 0

1) We need to convert 12.0 g of H2 into moles of H2, and 74.5 grams of CO into moles of CO
Molar mass of H2: M(H2) = 2*1.0= 2.0 g/mol
Molar mass of CO: M(CO) = 12.0 +16.0 = 28.0 g/mol

12.0 g H2 * 1 mol/2.0 g = 6.0 mol H2
74.5 g CO * 1 mol/28.0 g = 2.66 mol CO

2) Now we can use reaction to find out what substance will react completely, and what will be leftover.

CO + 2H2 -------> CH3OH
 1 mol 2 mol
given 2.66 mol 6 mol (excess)

How much
we need CO? 3 mol 6 mol

We see that H2 will be leftover, because for 6 moles H2 we need 3 moles CO, but we have only 2.66 mol CO.
So, CO will react completely, and we are going to use CO to find the mass of CH3OH.

3) CO + 2H2 -------> CH3OH
 1 mol 1 mol
 2.66 mol 2.66 mol

4) We have 2.66 mol CH3OH
Molar mass CH3OH : M(CH3OH) = 12.0 + 4*1.0 + 16.0 = 32.0 g/mol

2.66 mol CH3OH * 32.0 g CH3OH/ 1 mol CH3OH = 85.12 g CH3OH

Answer is D) 85.12 grams.

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If an atom has three shell and number of valence electrons are two. Is this element metal

cluponka [151]

Answer:

Metal

Element: Calcium

Valency: 2

Explanation:

To find the element, let's use the periodic table. (Look below)

We already went past 3 shells, just need the 2 electrons after it.

Just skip to the 4th row and count 2 to the right

We end up at Calcium.

Calcium is a metal and we're also on the alkaline earth metals column.

Calcium will need to lose 2 electrons to reach stability, so the valency is 2.

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3 years ago

When 7.085 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 21.71 grams of CO2 and 10.37 grams of H

Taya2010 [7]

Answer:

- Empirical:

$C_3H_7$

- Molecular:

$C_6H_{14}$

Explanation:

Hello,

In this case, based on the information regarding the combustion, the moles of carbon turn out:

$n_C=21.71gCO_2*\frac{1molCO_2}{44gCO_2}*\frac{1molC}{1molCO_2}=0.493molC$

Moreover, the moles of hydrogen:

$n_H=10.37gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{2molH}{1molH_2O}=1.152molH$

Thus, the subscripts of carbon and hydrogen in the hydrocarbon turn out:

$C=\frac{0.4934}{0.4934}=1\\H=\frac{1.15222}{0.4934}=2.335\\CH_{2.335}$

Now, looking for a suitable whole number we obtain the following empirical formula as 2.335 times 3 is 7 for hydrogen:

$C_3H_7$

In such a way, that compound has a molar mass of 43 g/mol, thus, the whole compound's molar mass is 86.18 g/mol for which the molecular formula is twice the empirical one, therefore:

$C_6H_{14}$

Which is hexane.

Best regards.

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3 years ago

Predict the boiling point of water at a pressure of 1.5 atm.

Lina20 [59]

Answer:

100.8 °C

Explanation:

The Clausius-clapeyron equation is:

$ln\frac{P_{1} }{P_{2}} =$ -Δ $\frac{H_{vap}}{r} (\frac{1}{T_{2}}-\frac{1}{T_{1}} )$

Where 'ΔHvap' is the enthalpy of vaporization; 'R' is the molar gas constant (8.314 j/mol); 'T1' is the temperature at the pressure 'P1' and 'T2' is the temperature at the pressure 'P2'

Isolating for T2 gives:

$T_{2}=(\frac{1}{T_{1}} -\frac{Rln\frac{P_{2}}{P_{1}} }{Delta H_{vap}}$

(sorry for 'deltaHvap' I can not input symbols into equations)

thus T2=100.8 °C

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3 years ago

For the following reaction, 33.7 grams of bromine are allowed to react with 13.0 grams of chlorine gas.

Ira Lisetskai [31]

Explanation:

The reaction is given as;

Br2(g) + Cl2(g) ----> 2BrCl(g)

From the equation;

1 mol of Br2 reacts with 1 mol of Cl2

Converting the masses given to moles, using the formular;

Number of moles = Mass / Molar mass

Br2;

Number of moles = 33.7 g / 159.808 g/mol = 0.21088 mol

Cl2;

Number of moles = 13.0 g / 70.906 g/mol = 0.18334 mol

From the values;

0.18334 mol of Cl2 would react with 0.18334 mol of Br2 with an excess of 0.02754 mol of Br2

What is the maximum amount of bromine monochloride that can be formed? __________grams

1 mol of Cl2 produces 2 mol of Bromine Monochloride

0.18334 mol of Cl2 would produce x

Solving for x;

x = 0.18334 * 2 = 0.36668 mol

Converting to mass;

Mass = Number of moles * Molar mass = 0.36668 mol * 115.357 g/mol

Mass = 42.299 g

What is the FORMULA for the limiting reagent?

The limiting reagent is Cl2 as it determines the amount of product formed. The moment the reaction uses up Cl2, the reaction stops.

What amount of the excess reagent remains after the reaction is complete? __________grams

The excess reagent is Br2

The number of moles left is;

0.02754 mol of Br2

Converting to mass;

Mass = Number of moles * Molar mass = 0.02754 mol * 159.808 g/mol

Mass = 4.401 g

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2 years ago

What happens to the glucose molecule during the process of cellular respiration?

Eduardwww [97]

Answer:

it's broken down into carbon dioxide and water

Explanation:

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