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3 years ago

If I start with 5 grams of c3h8 what is my theoretical yield of water

1 answer:

5 0

C3H8 (propane) undergoes combustion to produce water and carbon (IV) oxide according to equation below
C3H8(g) + 5 O2(g) ----> 3 CO2(g) + 4 H2O(l)

calculate the moles of C3H8 used
mole = mass/molar mass
molar mass of C3H8 = ( 12x3 )+( 1 x 8)=44 g/mol
moles of C3H8= 5g/ 44 g/mol= 0.114 moles
by use of mole ratio between C3H8 to H2O which is 1:4 therefore the mole of water= 0.114 x4=0.456 moles
mass of water =moles of water x molar mass of water

that is 0.456 moles x 18g/mol = 8.208 grams

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The decomposition of NI3 to form N2 and I2 releases −290.0 kJ of energy. The reaction can be represented as 2NI3(s)→N2(g)+3I2(g)

EastWind [94]

Answer:

-7.34 kilo Joules is the change in enthaply when 20.0 grams of nitrogen triiodide decomposes.

Explanation:

Mass of nitrogen triiodide = 20.0 g

Moles of nitrogen triiodide = $\frac{20.0 g}{395 g/mol}=0.05063 mol$

$2NI_3(s)\rightarrow N_2(g)+3I_2(g), \Delta H_{rxn}=-290.0 kJ$

According to reaction, 2 moles of nitrogen triiodide gives 290.0 kilo Joules of heat on decomposition ,then 0.05063 moles of nitrogen triiodide will give :

$\frac{-290.0 kJ}{2}\times 0.05063=-7.34 kJ$

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3 years ago

A 25.225 g sample of aqueous waste leaving a fertilizer manufacturer contains ammonia. The sample is diluted with 75.815 g of wa

Rudik [331]

Answer:

1.43 (w/w %)

Explanation:

HCl reacts with NH3 as follows:

HCl + NH3 → NH4+ + Cl-

1 mole of HCl reacts per mole of ammonia.

Mass of NH3 is obtained as follows:

Moles HCl:

0.02999L * (0.1068mol / L) = 3.203x10-3 moles HCl = Moles NH3

Mass NH3 in the aliquot:

3.203x10-3 moles NH3 * (17.031g / mol) = 0.0545g.

Mass of sample + water = 22.225g + 75.815g = 98.04g

Dilution factor: 98.04g / 14.842g = 6.6056

That means mass of NH3 in the sample is:

0.0545g * 6.6056 = 0.36g NH3

Weight percent is:

0.36g NH3 / 25.225g * 100

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