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Verizon [17]
2 years ago
9

What is the percent of hydrogen by mass in CH4O? (H:1.00 amu, C: 12.01 amu, O: 16.00 amu)

Chemistry
1 answer:
cricket20 [7]2 years ago
3 0

The mass percent of hydrogen in CH₄O is 12.5%.

<h3>What is the mass percent?</h3>

Mass percent is the mass of the element divided by the mass of the compound or solute.

  • Step 1: Calculate the mass of the compound.

mCH₄O = 1 mC + 4 mH + 1 mO = 1 (12.01 amu) + 4 (1.00 amu) + 1 (16.00 amu) = 32.01 amu

  • Step 2: Calculate the mass of hydrogen in the compound.

mH in mCH₄O = 4 mH = 4 (1.00 amu) = 4.00 amu

  • Step 3: Calculate the mass percent of hydrogen in the compound.

%H = (mH in mCH₄O / mCH₄O) × 100%

%H = 4.00 amu / 32.01 amu × 100% = 12.5%

The mass percent of hydrogen in CH₄O is 12.5%.

Learn more about mass percent here:brainly.com/question/4336659

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Various members of a class of compounds, alkenes, react with hydrogen to produce a corresponding alkane. Termed hydrogenation, t
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<u>Answer:</u> The mass of decane produced is 1.743\times 10^2g

<u>Explanation:</u>

To calculate the number of moles, we use the equation:  

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}       ......(1)

Mass of hydrogen gas = 2.45 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1:, we get:

\text{Moles of }H_2=\frac{2.45g}{2g/mol}=1.225mol

The chemical equation for the hydrogenation of decene follows:

C_{10}H_{20}(l)+H_2(g)\rightarrow C_{10}H_{22}(s)

As, decene is present in excess. So, it is considered as an excess reagent.

Thus, hydrogen gas is a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

1 mole of hydrogen gas produces 1 mole of decane.

So, 1.225 moles of hydrogen gas will produce = \frac{1}{1}\times 1.225=1.225mol of decane

Now, calculating the mass of decane by using equation 1, we get:

Moles of decane = 1.225 mol

Molar mass of decane = 142.30 g/mol

Putting values in equation 1, we get:

1.225mol=\frac{\text{Mass of decane}}{142.30g/mol}\\\\\text{Mass of carbon dioxide}=(1.225mol\times 142.30g/mol)=174.3g=1.743\times 10^2g

Hence, the mass of decane produced is 1.743\times 10^2g

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The percent composition by mass of nitrogen in NH OH (gram-formula mass = 35 grams/mole) is equal to
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3 years ago
Determine the number of grams of C4H10 that are required to completely react to produce 8.70 mol of CO2 according to the followi
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Answer:

126.4 g of C_{4}H_{10} are required

Explanation:

Balanced reaction: 2C_{4}H_{10}+13O_{2}\rightarrow 8CO_{2}+10H_{2}O

According to balanced reaction-

8 moles of CO_{2} are produced from 2 moles of C_{4}H_{10}

So, 8.70 moles of CO_{2} are produced from (\frac{2}{8}\times 8.70) moles of C_{4}H_{10} or 2.175 moles of C_{4}H_{10}

Molar mass of C_{4}H_{10} = 58.12 g/mol

So, mass of C_{4}H_{10} required = (2.175\times 58.12)g = 126.4 g

Hence 126.4 g of C_{4}H_{10} are required

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