Question

the amount of water produced by combustion of 286g of propane​

Answer 1

Answer:

How do I calculate the amount of water produced by the combustion of 16 g of methane?

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To solve such questions you must write a balanced reaction first.

CH4 (g)+2O2 (g)————-> CO2 (g) +2H2O (g)

Now according to the reaction 1 mole of CH4 gives 2 moles of H2O

And 16 grams of Methane mean 1 mole (moles=given mass/molecular mass. Molecular mass of methane is 16)

so 1 mole of methane = 16/16 (given mass 16 and molecular mass 16)

since 1 mole of methane is getting used up to form 2 mole of water.

number of moles=given mass/molecular mass ——- eq. 1

molecular mass of water is 18.

2=x/18 ( where x is the mass of water produced , using equation 1)

2*18=36grams

so 36 grams of water will be produced.

Answer 2

Answer:

286 grams of propane​ will produce 468 grams of water.

Explanation:

The reaction of propane with water proceeds as follows,

                       C₃H₈ + 5O₂ →  3CO₂ + 4H₂O                        (1)

Where,

C₃H₈ =propane

O₂=oxygen

CO₂=carbon dioxide

H₂O=water

The molar mass of propane is =44 grams

And the molar mass of water=18 grams

From equation (1) we can see that 44 grams of propane will produce 4×18 grams of water.

So, 1 gram of propane will produce,

$\frac{4\times 18}{44}=\frac{18}{11}$            (2)

Now 286 grams of propane will produce,

$\mathrm{\frac{18}{11}\times 286=468\ grams}$

So, 286 grams of propane​ will produce 468 grams of water.

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