[locagata48]

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

thanks for your help
I think that im doing something wrong with C dont 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) 9.5g Br * 1mol/80g Br * 22.4 L/ 1mol = 2.66 L Br

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

[Astrokel]

hey locagata48,

1A) The equation is wrongly balanced.
B) Check again after balancing
C) Check again

2A) hydrogen has and bromine exists as diatomic, check the rest after balancing

[Astrokel]

3

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

Check again

[macman104]

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

Oxygen gas is not 64g/mol.  Also, you can't compare masses when you calculate your percent yield, you need to calculate moles first, then do (moles actual/moles theoretical).

[Borek]

Nobody mentioned NH₄ yet?

[macman104]

Nobody mentioned NH₄ yet?

Lol, completely overlooked it.