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3 years ago

How are methane and propane alike

2 answers:

dsp733 years ago

8 0

Both are carbon chains made of just Carbon and Hydrogen. Both contain saturated bonds (sigma bonds). Both contain carbons that have spy hybridized orbitals. Both exhibit the combustible property of making carbon dioxide and water vapor when reacted (burned) with oxygen gas. Propane is a 3 carbon chain, while methane is just one carbon. Essentially, propane is just a longer version of methane, in that propane just includes more carbons.

saveliy_v [14]3 years ago

7 0

Both have Colorless gas, External MSDS, Phase Behavior solid, liquid, and gas.

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27.4 g of Aluminum nitrite and 169.9 g of ammonium chloride react to form aluminum chloride, nitrogen, and water. How many grams

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Answer: The mass of excess reagent (ammonium chloride) remained after the reaction is 62.7 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For aluminium nitrite:

Given mass of aluminium nitrite = 27.4 g

Molar mass of aluminium nitrite = 41 g/mol

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$\text{Moles of aluminium nitrite}=\frac{27.4g}{41g/mol}=0.668mol$

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Given mass of ammonium chloride = 169.9 g

Molar mass of ammonium chloride = 53.5 g/mol

Putting values in equation 1, we get:

$\text{Moles of ammonium chloride}=\frac{169.9g}{53.5g/mol}=3.176mol$

The chemical equation for the reaction of aluminium nitrite and ammonium chloride follows:

$Al(NO_2)_3+3NH_4Cl\rightarrow AlCl_3+3N_2+6H_2O$

By Stoichiometry of the reaction:

1 mole of aluminium nitrite reacts with 3 moles of ammonium chloride

So, 0.668 moles of aluminium nitrite will react with = $\frac{3}{1}\times 0.668=2.004mol$ of ammonium chloride.

As, given amount of ammonium chloride is more than the required amount. So, it is considered as an excess reagent.

Thus, aluminium nitrite is considered as a limiting reagent because it limits the formation of product.

Excess moles of ammonium chloride = (3.176 - 2.004) mol = 1.172 moles

Calculating the mass of ammonium chloride by using equation 1, we get:

Excess moles of ammonium chloride = 1.172 moles

Molar mass of ammonium chloride = 53.5 g/mol

Putting values in equation 1, we get:

$1.172mol=\frac{\text{Mass of ammonium chloride}}{53.5g/mol}\\\\\text{Mass of ammonium chloride}=(1.172mol\times 53.5g/mol)=62.7g$

Hence, the mass of excess reagent (ammonium chloride) remained after the reaction is 62.7 grams

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