From the information below, determine the mass of substance C that is formed if

Question

From the information below, determine the mass of substance C that is formed if 45.0 g of substance A reacts with 23.0 g of substance B, assuming that the reaction between A and B goes to completion.

A. Substance A is a solid that consists of a Group 13 (3A) metal (boron is not a metal!) and carbon; the m% C in substance A is 57.2%. It reacts with substance B to form substances C and D. 4.0 x 1016 formula units of substance A weighs 8.371 micrograms. Show all work

B. 47.9 g of substance B contains 5.36 g H and 42.5 g O. Show all work.

C. When 10.0 g of substance C is burned in excess oxygen, 33.8 g of CO2and 6.92 g of H2O are produced. The molar mass of substance C is MC = ~26 g/mol.

D. Substance D is the hydroxide of the metal in substance A. Show all work.

Explanation / Answer

A. find moles of substance A = 4 x 10^16 formula units/6.022 x 10^23 = 6.64 x 10^-8 mols

mass of A = 8.371 micrograms = 8.37 x 10^-6 g

molar mass of substance A = 8.37 x 10^-6/6.64 x 10^-8 = 126.05 g/mol

m% C in substance A = 57.2%

mass in g of C in 1 mole of substance A = 0.572 x 126.05 = 72.101 g

moles of C = 72.101/12.01 = 6.0 mols

So we have 6 moles of C in 1 mole of substance A

molar mass of metal in substance A = 126.05 - 72.101 = 53.95 g

The metal is thus Al,

The molecular formula of substance A= Al2C6

B. 47.9 g of substane B gave 5.36 g of H and 42.5 g of O

moles of H = 5.36/1.01 = 5.54 mol

moles of O = 42.5/16 = 2.66 mol

divide by smallest factor,

H = 5.54/2.66 = 2

O = 5.54/5.54 = 1

So we have molecular formula of substance B = H2O

C. mols of C in 33.8 g CO2 = 33.8/44.01 = 0.77 mol

mols of H in 6.92 g of H2O = 2 x (6.92/18.015) = 0.77 mol

Ratio being 1 : 1

So emiprical formula becomes CH

empirical formula mass = 12 + 1 = 13

molar mass = 26

factor = 26/13 = 2

molecular formula of substance C = C2H2

D. Is the hydroxide of metal A

molecular formula of substance D = Al(OH)3

Now when 45.0 g of A is reacted with 23.0 g of B

Al2C6 + 6H2O ---> 3C2H2 + 2Al(OH)3

moles of A = 45/126.05 = 0.36 mols

moles of B = 23/18.015 = 1.28 mols

If all of Al2C6 has reacted we would require = 6 x 0.36 = 2.16 mols of B

If all of B has reacted we would require = 1.28/6 = 0.21 mols of A

Since we have excess of A then required, H2O is the limiting reagent here

So mass of substance C formed = 1.28 x 0.5 x 26 = 16.64 g

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