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Ronch [10]
3 years ago
9

What mass of oxygen is needed for the complete combustion of 4.60×10−3g of methane?

Chemistry
1 answer:
sp2606 [1]3 years ago
7 0
First step in answering the question is to establish a balanced chemical reaction equation. More specifically, a combustion chemical equation. 

CH4 + 2O2 ---> CO2 + 2H20

Then using dimension analysis: 

4.60*10^{-3} g CH4 ( \frac{moleCH4}{16 g CH4}) * ( \frac{2mole O2}{mole CH4}) * ( \frac{32 g O2}{mole O2} ) =  0.0184 g O_{2}
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Explanation:

I hope this helps

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This is why the third option is the correct one.

4 0
3 years ago
The number of ions in 17.1 gm of aluminum sulphate Al2 (SO4)3 =….. [Molar mass of Al2 (SO4)3 = 342 gm]
katen-ka-za [31]

Answer:

1.51 X 10^23 ions

Explanation:

The number of ions in 17.1 gm of aluminum sulphate Al2 (SO4)3 =….. [Molar mass of Al2 (SO4)3 = 342 gm]

in one molecule of Al2(SO4)3 there are 5 ions 2 aluminum and 3 sulfate ions

in 2 molecules there are 2X5= 10 ions

in 10 molecules there are 10X5 = 50 ions

molar mass of Al2(SO4)3 = (2 X 26.98) +( 3 X 32.1) + (3 X 4 X 16.0 ) =342.gms = 17.1/342 =0.0500 moles

1  mole =6.02 X 10^23 molecules ( see Avogadros number)

0.0500 moles = 0.0500 X 6.02 X 10^23 molecules =

0.301 X 10^23 molecules = 3.01 X 10^22 molecules

We determined that each molecule of Al2(SO4)3 has 5 ions

so 3.01 X10^22 molecules have 5 X 3.01 X 10^22 ions =

15.05 X 10^22 ions = 1.51 X 10^23 ions

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