Chemistry Homework; help!?

How many grams of carbon dioxide are produced from the combustion of 2.7 moles of ethane in the presence of 10 moles of oxygen?

C2H6 + 7O2 → 4CO2 + 6H2O

A) 62 G

B) 45 G

C) 250 G

D) 120 G

Update:

How many grams of ammonia are produced when 54.0 g of water react with 123 g of magnesium nitride?Mg3N2(s) 6H2O(l) → 2NH3(aq) 3Mg(OH)2(s)

(A) 41.4 g

(B) 10.3 g

(C) 34.0 g

(D) 17.0 g

Update 2:

What mass of copper metal is produced when 12.7 grams of aluminum react with 8.47 grams of copper(II) chloride?3CuCl2 2Al → 2AlCl3 3Cu

(A) 0.109 g

(B) 4.00 g

(C) 1.33 g

(D) 43.8 g

Update 3:

Which quantity is not conserved in the reaction below?

CH4(g) Br2(g) → CH3Br(g) HBr(g)

(A) mass

(B) formula units

(C) moles

(D) number of atoms

1 Answer

Relevance
  • Lexi R
    Lv 7
    6 years ago
    Favorite Answer

    How many grams of carbon dioxide are produced from the combustion of 2.7 moles of ethane in the presence of 10 moles of oxygen?

    C2H6 + 7O2 → 4CO2 + 6H2O

    O2 is the limiting reagent

    mass CO2 = 10 mol O2 x (4 mol CO2 / 7 mol O2) x (44.01 g CO2 / 1 mol CO2)

    = 250 g

    How many grams of ammonia are produced when 54.0 g of water react with 123 g of magnesium nitride?Mg3N2(s) 6H2O(l) → 2NH3(aq) 3Mg(OH)2(s)

    work out how much NH3 you would get from complete reaction of all of each reagent. the one that produces the least NH3 is the limiting reagent and that amount of NH3 is the amount of NH3 possible

    g NH3 = 54.0 g x H2O x (1 mol H2O / 18.016 g) x (2 mol NH3 / 6 mol H2O) x (17.034 g NH3 / 1 mol)

    = 17.0 g

    g NH3 = 123 g MgN3 x (1 mol MgN3 / 66.34 g MgN3) x (2 mol NH3 / 1 mol MgN3) x (17.034 g NH3 / 1 mol NH3)

    = 63.2 g

    you can get 17.0 g NH3

    What mass of copper metal is produced when 12.7 grams of aluminum react with 8.47 grams of copper(II) chloride?3CuCl2 2Al → 2AlCl3 3Cu

    mass Cu = 12.7 g Al x (1 mol Al / 26.98 g Al) x (3 mol Cu / 2 mol Al) x (63.55 g Cu / 1 mol Cu)

    = 44.9 g

    mass Cu = 8.47 g CuCl2 x ( 1 mol CuCl2 / 134.45 g CuCl2) x (3 mol Cu / 3 mol CuCl2) x (63.55 g Cu / 1 mol Cu)

    = 4.00 g

    CuCl2 is the limiting reagent, as the amount provided produces the least Cu, 4.00 g of Cu will be produced

    (B) 4.00 g

    Which quantity is not conserved in the reaction below?

    CH4(g) Br2(g) → CH3Br(g) HBr(g)

    what a ridiculous question

    the number of atoms is the same on each side of the arrow, so atoms are conserved

    if the number of atoms on each side is constant then the mass must be as well

    there are 2 moles of gas on the LHS of the arrow and 2 moles on the RHS, so moles are conserved

    So by default the answer must be formula units... but that is because there are not any formula units in the reaction, formula units are what we call the smallest units of ionic compounds, and there are no ionic compounds in this reaction, they are all covalent compounds. So if you wanted to get technical about it the number of formula units is constant as well .... zero.

    http://en.wikipedia.org/wiki/Formula_unit

    • Login to reply the answers
Still have questions? Get your answers by asking now.