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antoniya [11.8K]
3 years ago
5

Help will mark brainlest!

Chemistry
1 answer:
vladimir2022 [97]3 years ago
7 0
D I just took this and got it right
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Force /Area = Pressure
Force = Pressure * Area
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I hope this helps!
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Determine the limiting reactant when 30.0 g of propane, C3H8, is burned with 75.0 g of oxygen.
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<u>Answer:</u> The limiting reagent is oxygen gas.

<u>Explanation:</u>

Limiting reagent is defined as the reactant that is present in less amount and it limits the formation of products.

Excess reagent is defined as the reactant which is present in large amount.

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

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

  • <u>For propane:</u>

Given mass of propane = 30.0 g

Molar mass of propane = 44.1 g/mol

Putting values in equation 1, we get:

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  • <u>For oxygen:</u>

Given mass of oxygen = 75.0 g

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Putting values in equation 1, we get:

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The chemical equation for the combustion of propane follows:

C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(l)

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5 moles of oxygen gas reacts with 1 mole of propane.

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

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

Hence, the limiting reagent is oxygen gas.

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