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3 years ago

A sample of a compound contains 30.0 g C and 5.05 g H. Its molar mass is 78.12 g/mol. What is the compound’s molecular formula?

1 answer:

8 0

You seems to look worried so i will answer your question.
The correct answer is C6H6 as it has a molar mass of 73.07 grams per mole and it is the sample of compound that sums up with 78.12 grams per mole.
CH has 13.02 grams per mole.C2H2 has a molar mass of 26.04.

Hope this helps you : )
If have more questions feel free to ask me :D

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melamori03 [73]

Answer: The mass of nickel (II) oxide and aluminium that must be used is 18.8 g and 4.54 g respectively.

Explanation:

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

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

For nickel:

Given mass of nickel = 14.8 g

Molar mass of nickel = 58.7 g/mol

Putting values in equation 1, we get:

$\text{Moles of nickel}=\frac{14.8g}{58.7g/mol}=0.252mol$

For the given chemical reaction:

$3NiO(s)+2Al(s)\rightarrow 3Ni(l)+Al_2O_3(s)$

For nickel (II) oxide:

By Stoichiometry of the reaction:

3 moles of nickel are produced from 3 moles of nickel (II) oxide

So, 0.252 moles of nickel will be produced from $\frac{3}{3}\times 0.252=0.252mol$ of nickel (II) oxide

Now, calculating the mass of nickel (II) oxide by using equation 1:

Molar mass of nickel (II) oxide = 74.7 g/mol

Moles of nickel (II) oxide = 0.252 moles

Putting values in equation 1, we get:

$0.252mol=\frac{\text{Mass of nickel (II) oxide}}{74.7g/mol}\\\\\text{Mass of nickel (II) oxide}=(0.252mol\times 74.7g/mol)=18.8g$

For aluminium:

By Stoichiometry of the reaction:

3 moles of nickel are produced from 2 moles of aluminium

So, 0.252 moles of nickel will be produced from $\frac{2}{3}\times 0.252=0.168mol$ of aluminium

Now, calculating the mass of aluminium by using equation 1:

Molar mass of aluminium = 27 g/mol

Moles of aluminium = 0.168 moles

Putting values in equation 1, we get:

$0.168mol=\frac{\text{Mass of aluminium}}{27g/mol}\\\\\text{Mass of aluminium}=(0.168mol\times 27g/mol)=4.54g$

Hence, the mass of nickel (II) oxide and aluminium that must be used is 18.8 g and 4.54 g respectively.

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3 years ago

A 1.42-g sample of a pure compound with formula m2so4 was dissolved in water and treated with an ex- cess of aqueous calcium chl

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The reaction between $m_{2}SO_{4}$ with calcium chloride can be shown as- $m_{2}SO_{4}$ + $CaCl_{2}$ → $CaSO_{4}$ ↓+2mCl. The molecular weight of $CaSO_{4}$ is 136.14g. The weight of sulfate ion is 96.06g. The molecular weight of $m_{2}SO_{4}$ = (2×m + 96.06). From the reaction we can see that 1 mole of calcium chloride reacts with 1 moles $m_{2}SO_{4}$ to produce 1 mole of calcium sulfate. Now 1.36g of calcium sulfate is equivalent to 1.36/136.14=9.989× $10^{-3}$ moles of calcium sulfate.

Thus, 9.989× $10^{-3}$ moles of $m_{2}SO_{4}$ reacts in this reaction.

Let assume the atomic mass of m is x thus the molecular weight of $m_{2}SO_{4}$ is 2x+96.

So we may write 9.989× $10^{-3}$ × (2x+96) =1.42

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Or, 2x = 46.146

Or, x = 23.073

Thus the atomic mass of m is 23.073. The atom (m) is sodium (Na).

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