# When propane gas, C3H8, is used as a fuel for many barbecuesâ€¦?

Can you please show me step by step on how to get these answers?

How many grams of the compound are in 1.50 mol propane?

How many moles of the compound are in 34.0 g propane?

How many grams of carbon are in 34.0 g propane?

How many atoms of hydrogen are in 0.254 g propane?

Molar mass of propane is (3 C X 12 = 36 g) + (8 H X 1 = 8 g), thus it is 44 g/mol.

1.50 mol propane X (44 g/mol) = 66 g propane

34.0 g propane (1 mol/44 g) = 0.77 mol propane

In C3H8 there are 3 mol C per mol of propane, so since we determined that 34.0 g propane are 0.77 mol propane, we can do this:

0.77 mol propane (3 mol C/1 mol propane) = 2.3 mol C

0.254 g propane (1 mol/44 g) = 0.0058 mol propane.

Now ratio is 8 mol H per mol of propane

0.0058 mol propane (8 mol H/1 mole propane) = 0.046 mol H.

Avogadro's number ==> 1 mol = 6.022 X 10^23

0.046 mol H (6.022 X 10^23/mol H) = 2.8 X 10^21 atoms H

1. look at a periodic table-add 3 carbons and 8 hydrogens to get molar mass. multiply by 1.5.

2. divide 34g by molar mass to get moles of propane

3. grams of carbon is simply the ratio of carbons so you take the mass of carbons (~36) divided by molar mass to get ratio. then multiply that ratio by 34g.

4. one mole=6.022x10e23 molecules. So there are 6.022x10e23 molecules of propane in the molar mass (~44). divide .254g/molar mass to get moles propane. then multiply by 6.022x10e23 to get number of molecules propane. there are 8 atoms of H for every molecule so multiply 8 by number of propane molecules

be sure to look up the real values of mass in periodic table, my answers were estimates

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How many grams of the compound are in 1.50 mol propane?

How many moles of the compound are in 34.0 g propane?

How many grams of carbon are in 34.0 g propane?

How many atoms of hydrogen are in 0.254 g propane?

**Answer:**Molar mass of propane is (3 C X 12 = 36 g) + (8 H X 1 = 8 g), thus it is 44 g/mol.

1.50 mol propane X (44 g/mol) = 66 g propane

34.0 g propane (1 mol/44 g) = 0.77 mol propane

In C3H8 there are 3 mol C per mol of propane, so since we determined that 34.0 g propane are 0.77 mol propane, we can do this:

0.77 mol propane (3 mol C/1 mol propane) = 2.3 mol C

0.254 g propane (1 mol/44 g) = 0.0058 mol propane.

Now ratio is 8 mol H per mol of propane

0.0058 mol propane (8 mol H/1 mole propane) = 0.046 mol H.

Avogadro's number ==> 1 mol = 6.022 X 10^23

0.046 mol H (6.022 X 10^23/mol H) = 2.8 X 10^21 atoms H

1. look at a periodic table-add 3 carbons and 8 hydrogens to get molar mass. multiply by 1.5.

2. divide 34g by molar mass to get moles of propane

3. grams of carbon is simply the ratio of carbons so you take the mass of carbons (~36) divided by molar mass to get ratio. then multiply that ratio by 34g.

4. one mole=6.022x10e23 molecules. So there are 6.022x10e23 molecules of propane in the molar mass (~44). divide .254g/molar mass to get moles propane. then multiply by 6.022x10e23 to get number of molecules propane. there are 8 atoms of H for every molecule so multiply 8 by number of propane molecules

be sure to look up the real values of mass in periodic table, my answers were estimates

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