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ivanzaharov [21]
3 years ago
14

The equation below shows the decomposition of lead nitrate. how many grams of oxygen are produced when 11.5g NO2 is formed?

Chemistry
1 answer:
Soloha48 [4]3 years ago
7 0


Number of moles of NO2 = 11.5/(14+16x2) =0.250mol
Mole ratio: 4NO2 to 1O2
Number of moles of O2=0.250/4 =0.0625mol
Mass of O2 produced= 0.0625x32 =2g
Answer: B
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Determine the [h3o+] of a 0.210 m solution of formic acid.
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When the Pka for formic acid = 3.77
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3 0
3 years ago
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!

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