Propane (C3H8) is widely used in liquid form as a fuel for barbecue grills and camp stoves. For 67.7 g of propane, determine the following.(a) Calculate the moles of compound. mol(b) Calculate the grams of carbon. g

A. 1.54 mole.

B. 55.39g of carbon

Explanation:

A. Determination of the number of mole in 67.7g of C3H8.

Mass of C3H8 = 67.7g

Molar mass of C3H8 = (3x12) + (8x1) = 44g/mol

Number of mole of C3H8 =..?

Number of mole = Mass/Molar Mass

Number of mole of C3H8 = 67.7/44

Number of mole of C3H8 = 1.54 mole

B. Determination of the mass of carbon in the compound.

This is illustrated below:

The mass of C in compound can be obtained as follow:

=> 3C/C3H8 x 67. 7

=> 3x12 / 44 x 67.7

=> 36/44 x 67.7

=> 55.39g

Therefore, 55.39g of carbon is present in the compound.

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2