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Andrew [12]
1 year ago
14

40.0L of N₂ gas are in a sealed container at STP.How many moles of N₂ are present?9 mol

Chemistry
1 answer:
Vinvika [58]1 year ago
8 0

Explanation:

We have to find the number of moles of N₂ that are present in a sample that has a volume of 40.0 L at STP.

STP means Standard Conditions of Temperature and Pressure. These conditions are 273.15 K and 1 atm. We know that 1 mol of N₂ will occupy 22.4 L. We can use that ratio to find the answer to our problem.

1 mol of N₂ = 22.4 L

moles of N₂ = 40.0 L * 1 mol/(22.4 L)

moles of N₂ = 1.79 mol

Answer: 1.79 moles of nitrogen are present.

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How do the strength and the cohesion of clay-rich regolith or soil change with the addition of water?
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Which statement accurately describes the charge of the nucleus of an atom? (2 points) The nucleus can be either positively charg
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5 0
3 years ago
Read 2 more answers
1.12g H2 is allowed to react with 9.60 g N2, producing 1.23 g NH3.
andriy [413]

Answer:

A. m_{NH_3}^{theo} =1.50gNH_3

B. Y=82.2\%

Explanation:

Hello!

In this case, since the undergoing chemical reaction between nitrogen and hydrogen is:

N_2+3H_2\rightarrow 2NH_3

Thus we proceed as follows:

A. Here, we first need to compute the moles of ammonia yielded by each reactant, in order to identify the limiting one:

n_{NH_3}^{by \ H_2}=1.12gH_2*\frac{1molH_2}{2.02gH_2}*\frac{2molNH_3}{3molH_2}=0.370molNH_3\\\\  n_{NH_3}^{by \ N_2}=1.23gN_2*\frac{1molN_2}{28.02gN_2}*\frac{2molNH_3}{1molN_2}=0.0878molNH_3

Thus, since nitrogen yields the fewest moles of ammonia, we realize it is the limiting reactant, so the theoretical yield, in grams, of ammonia is:

m_{NH_3}^{theo}=0.0878mol*\frac{17.04gNH_3}{1molNH_3} =1.50gNH_3

B. Finally, since the actual yield of ammonia is 1.23, the percent yield turns out:

Y=\frac{1.23gNH_3}{1.50gNH_3} *100\%\\\\Y=82.2\%

Best regards!

5 0
3 years ago
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