In one reaction with a 90 - yield- 200 mL of 2 M N a 2 CO 3 aq - 27 g NO and a large excess O 2 are allowed to react- 2 N a 2 CO 3 aq - 4 NO g - O 2 g - 4 N a N O 2 aq -2 CO 2 g what mass of N aNO 2 is obtained based on experimental yield-A- 23-29 gB- None of theseC- 49-68 gD- 46-58 g

The correct option is C 49.68 gNumber of moles of Na_2CO_3 nbsp; =volume (L)×molarity = 200 × 21000 nbsp; = 0.4 mol Number of moles of NO = 2730 = 0.9 m ...

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Standard XII

Chemistry

Equivalent Mass

In one reacti...

Question

In one reaction with a 90% yield, 200 mL of 2 M $N a_{2} C O_{3} (a q), 27 g N O$ and a large excess $O_{2}$ are allowed to react. $2 N a_{2} C O_{3} (a q) + 4 N O (g) + O_{2} (g) \to 4 N a N O_{2} (a q) + 2 C O_{2} (g)$ what mass of $N a N O_{2}$ is obtained based on experimental yield?

49.68 g

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46.58 g

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23.29 g

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None of these

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Solution

The correct option is A 49.68 g
Number of moles of $N a_{2} C O_{3}$
$= volume (L) \times molarity = \frac{200 \times 2}{1000} = 0.4 mol$

Number of moles of NO $= \frac{27}{30} = 0.9 mol$

Finding the limiting reagent:
$\frac{moles of N a_{2} C O_{3}}{2} = \frac{0.4}{2} = 0.2$
$\frac{moles of N O}{4} = \frac{0.9}{4} = 0.225$
Hence, $N a_{2} C O_{3}$ is the limiting reagent.
2 mol $N a_{2} C O_{3}$ gives 4 moles $N a N O_{2}$
0.4 mol $N a_{2} C O_{3}$ gives
$= 2 \times 0.4 = 0.8 mol N a N O_{2}$
$= 0.8 \times 69 g N a N O_{2}$
$= 55.2 g N a N O_{2}$ with 100% yield
$= 55.2 \times 0.90 g N a N O_{2}$ with 90% yield
$= 49.68$ g

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In one reaction with a 90% yield, 200 mL of 2 M Na2CO3(aq),27 g NO and a large excess O2 are allowed to react. 2Na2CO3(aq)+4NO(g)+O2(g)→4NaNO2(aq)+2CO2(g) what mass of NaNO2 is obtained based on experimental yield?

In one reaction with a 90% yield, 200 mL of 2 M $N a_{2} C O_{3} (a q), 27 g N O$ and a large excess $O_{2}$ are allowed to react. $2 N a_{2} C O_{3} (a q) + 4 N O (g) + O_{2} (g) \to 4 N a N O_{2} (a q) + 2 C O_{2} (g)$ what mass of $N a N O_{2}$ is obtained based on experimental yield?