From the information below,determine the mass of C that will be formed if 45.0 g

Question

From the information below,determine the mass of C that will be formed if 45.0 grams of substance A reacts with 23.0 grams of substance B. Assume that the reaction between A and B goes to completion.) a. Substance A is a gray solid that consists of an alkaline earth metal and carbon (37.5% by mass). It reacts with substance B to produce substances C and D.Forty m on trillion formula units of A have a mass of 4.26 milligrams. Find the molar mass of substance A, then find mass of Cin substance A Finally find molar mass of Group ll ement and determine formula of substance A. b. 47.9 grams of substance B contains 5.36 grams of hydrogen and 42.5 grams of oxygen. Find empirical formula of B.(What assumption are you going to make about c. When 10.00 grams of c is burned is burned in excess oxygen,33.8 grams of carbon dioxide and 6.92 grams of water are produced. A mass spectrum of substance C shows a parent molecular ion with a charge-to-mass ratio of 26 (i.e. molar mass is 26 g/mol) Use combustion analysis (SEE Chapter 4) to determine empirical formula and molecular formula of C d. Substance D is the hydroxide of the metal in substance A. Balance the equation now using A+B- C +D Find the limiting reactant and then determine grams of C produced

Explanation / Answer

a). 40 million trillion molecules having a mass 4.26 mg

40 x 1018

Number of moles present in this x 1018/Avogadro number = 40 x 1018/6.02 x 1023 = 6.64 x10-5

Mass of  6.64 x10-5 moles is 4.26 x10-3 g or

mass of one mole = (4.26 x10-3 /6.64 x10-5 ) =0.641 102 = 64.1 g

Molecular mass of compound is the mass of one mole of substance = 64.1 g/mole.

Given that it contains C as 37.5 % and thus the compound is CaC2 (A) (Ca 40 and C 12 + 12 = 14)

Without knowing about B calculating C is difficult.

Hb). 47.9 g of B contains 5.36 g H and 42.5 g O.

% of H = 5.36/47.9 = 0.111 (Atomic weight 1)

% of O = 42.5/47.9 = 0.888 (Atomic weight 16)

Dividing both of them with their atomic weight

H will give 0.111

O will give = 0.0555

Divide with minimum (with 0.0555)

H will come as 2

and O as 1

Thus empirical formula is H2O, which is equal to the molecular formula.

c).

10 g C gives 33.8 g of CO2 and 6.92 g water on burning.

From mass spectra, the molecular mass of C is 26.

Thus 10 g means

10/26 = 0.384 moles

similarly for CO2

number of moles = 33.8/44 = 0.76 moles

number of mole of water is = 6.92/18 = 0.38 moles.

Thus The molecule contains two C and 2 H (From mole ratio of CO2 and C will come as 2. similarly one mole C produces one-mole water. Note one-mole water contains 2 mole H) ie C2H2

d). The reaction is

2CaC2 (s) + 2 H2O (l) -----> C2H2 (g) + Ca(OH)2 (aq)

45 g CaC2 contains = 45/64.1 = 0.70 moles

23 g water contains = 23/18 = 1.27 moles

As per the balanced equation, 2 mole CaC2 reacts with 2-mole water. Thus limiting reagent is A (The one in less amount)

Thus number of moles of C formed will be the half of the number of moles of A = 0.35 moles.

Mass in g = 0.35 * 26 = 9.1 g

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