From the information below, determine the mass of substance C that will be forme

Question

From the information below, determine the mass of substance 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 substance C and D. Forty million trillion formula usnits of A have a mass of 4.26 milligrams.

b) 47.9 grams of substance B contains 5.36 grams of hydrogen and 42.5 grams of oxygen.

c) when 10.0 grams of C is burned in excess oxygen, 33.8 grams of carbon dioxide and 7.92 grams of water are produced. A mass spectrum of substance C shows a parent molecular ion with a mass-to-charge ratio of 26.

d) Substance D is the hydroxide of the metal in substance A.

Explanation / Answer

typically you start with these steps

1) write a balanced equation
2) convert mass to moles
3) determine limiting reagent
4) convert moles limiting to moles other
5) convert moles back to mass ... this is theoretical mass
6) % yield if needed = actual mass / theoretical mass x 100%

keeping that in mind.. your problem....

*** step 1... balanced equation.***

n A + m B -----> p C + q D.. something like that right?

so we need to figure out what A, B, C, and D are first...

*** Substance A ***.

A is of the form MexCy... where Me is metal

and you were given....
40,000,000 x 1 trillion = 4x10^7 x 10^12 = 4x10^19 units = 4.26x10^-3 g... translated...

molar mass = 4.26x10^-3 g / 4x10^19 units x (6.022x10^23 units / mole) = 64.1 g/mole

of that 37.5% is C by mass and 62.5% is Me...

1 mole would have
carbon = 0.375 x 64.1g = 24.0 g x (1 mole / 12g) = 2 moles C
Me = 0.625 x 64.1g = 40.1 g Me

For 1 mole MexC2.. X must be a whole #..
so for 1 mole to have 40.1 g Me
40.1 g / atomic mass of Me must = an integer...

From a periodic table, the alkaline earths are Be (mass 9.0), Mg (mass 24.3), Ca (mass 40.1).. and so on.. the only element that will give a whole # is Calcium...so A is CaC2... calcium carbide

*** substance B ***
since 47.9 grams substance contains 5.36 g H and 42.5 g O
and since = 5.36 g + 42.5 g = 47.9 g
the substance is made entirely of H and O...

moles H = 5.36 g x (1 mole / 1 g) = 5.36 moles
moles O = 42.5 g x (1 mole / 16 g) = 2.66 moles

divide by the smallest 2.66

5.36 / 2.66 = 2
2.66 / 2.66 = 1

so the empirical formula of B is H2O... which means B is H2O in this case... We won't have H4O2, H6O3.. etc..

*** substance C ***
the idea here is this... all of the C in CO2 and all of the H in H2O came from the compound.. none of it came from the O2 we reacted it with right?.. so let's find moles and mass of C and H in the compound...

moles CO2 = 33.8 g x (1 mole / 44.0g) = 0.768 moles CO2
since 1 mole CO2 contains 1 mole C
moles C = 0.768

moles H2O = 6.9 g x (1 mole / 18.0 g) = 0.383
since 1 mole H2O contains 2 moles H
moles H = 0.383 moles H2O x (2 moles H / 1 mole H2O) = 0.767

mass C = 0.768 x 12 = 9.22 g
mass H = 0.767 x 1 = 0.767 g

since mass C + mass H = 10.0 g
and since we started with 10.0 g of substance C
substance C is therefore made entirely of Carbon and Hydrogen

and the mole ratio of C:H = 0.768:0.767 = 1:1.. so empirical formula is CH...

empirical unit mass of CH = 12+1 = 13
parent ion mass = 26..

so there are 26/13 = 2 units of CH in each molecule.. ie substance C is C2H2

*** substance D ***
hydroxide of Ca is Ca(OH)2

so the balanced reaction is

CaC2 + 2 H2O ----> C2H2 + Ca(OH)2

*** step 2 ***
moles CaC2 = 45.0 g x (1 mole / 64.1 g) = 0.702 moles
moles H2O = 23.0 g x (1 mole / 18.0 g) = 1.278 moles

*** step 3 ***
from the balanced equation.. 1 mole CaC2 reacts with 2 moles H2O..so...

0.702 moles CaC2 x (2 moles H2O / 1 mole CaC2) = 1.404 moles H2O

meaning that is amount of H2O required to react with all the CaC2...since we only have 1.278 moles H2O available, we don't have enough water to completely consume all the CaC2 we started with. H2O is therefore the limiting reagent

*** step 4 ***
from the balance equation, 2 moles H2O ---> 1 mole C2H2...so...

1.278 moles H2O x (1 moles C2H2 / 2 moles H2O) = 0.639 moles C2H2

*** step 5 ***
0.639 moles C2H2 x (26.0 g / mole) = 16.6 g C2H2

********************
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