1) When ammonia gas, oxygen gas and methane gas (CH4) are combined, the products are hydrogen cyanide gas and water.
A) write a balance equation
B) calculate the mass of each product produced when 225 g of oxygen gas is reacted with an excess of the other two reactants
C) if the actual yield of the experiment in B) is 105 g of HCN, calculate the percent yield

A) NH4 + 4O2 + CH4 -> HCN + 4H2O
B) 225g O * 1 mol/64 g * 1 mol HCN/ 4 mol O * 27 g/ 1mol = 23.73g HCN
225g O * 1mol/64 g * 4 mol/4 mol * 18 g/1mol = 63.28 g HCN
C) 105 g HCN/ 23.73 g *100% = 442.47
I think that I'm doing something wrong with C don't think that it should be that high


2) Hydrogen gas and bromine gas react to form hydrogen bromide gas
A) write balanced equation
B) how many grams of hydrogen bromide gas can be produced from 3.2 g of hydrogen gas and 9.5 g of bromine gas
C) how many grams of which reactant is left unreacted?
D) what volume of HBr, measured at STP, is produced in B)

A) H + Br -> HBr
B) 3.2 g H * 1 mol/ 1 g H * 1mol/ 1mol * 81g HBr/1 mol = 259.2g HBr
9.5g Br * 1mol/ 80g Br * 1mol/1mol * 81g HBr/1mol = 9.62g HBr
C)9.5-9.62 = .12
D)i don't get this one


3) Fe2O3 + 2Al -> 2Fe + Al2O3
what mases of Fe2O3 and Al must be used to produce 15.0g Fe? What is the maximum mass of the Al2O3 that could be produced

15.0g Fe*1mol/112g * 2 mol.1mol * 160/1mol = 42.86g Fe2O3
15.0g Fe * 1mol/112g * 2mol/2 mol * 45g Al/1mol = 7.232g Al

42.86+7.232 = 15.0+x
50.092 = 15.0x
50.092/15.0 = 3.34g Al

1) Check out your balanced equation: I find more H atoms on the RHS and less O atoms on the same hand side.

2) Remember, hydrogen and bromine are diatomic molecules, so, H2 and Br2

3) I need to do it on paper and have the periodic table with me now (which is not the case) Will see to that later, promise.

Why KCL has a higher melting point than I

Nope for the Fe2O3. It's half that value, since 1 mol of Fe2O3 produces 2 mol of Fe.
Good for Al.
I never found such a method to find the mass of Al2O3 in the above experiment... I got 13.66g.

mygabriel, start another thread, please. Will explain that there, I you wish to have some ideas about your question.