X g of ethylamine is subjected to reaction with followed by water; evolved dinitrogen gas which occupied 2.24 L volume at STP. X is ______ g.

Question

X g of ethylamine is subjected to reaction with ${NaNO}_{2} / HCl$ followed by water; evolved dinitrogen gas which occupied 2.24 L volume at STP. X is ______ $\times 10^{- 1}$ g. [2024]

Smart Achievers · Accepted Answer

(45)

$C_{2} H_{5} {NH}_{2} \overset{{NaNO}_{2} + HCl}{\to} C_{2} H_{5} N_{2}^{+}$

$\overset{H_{2} O}{\to} C_{2} H_{5} OH + N_{2}$

Moles of nitrogen

$(n_{N_{2}}) = \frac{Volume at STP in L}{22.4} = \frac{2.24}{22.4} mol = 0.1 mol$

From reaction it is clear that moles of ethylamine $(n_{C_{2} H_{5} {NH}_{2}}) = n_{N_{2}} = 0.1 mol .$

Mass of ethylamine = $n_{C_{2} H_{5} {NH}_{2}} \times$ molar mass of ethylamine

$= 0.1 mol \times 45 {g mol}^{- 1} = 45 \times 10^{- 1} g$

Solution

Q.
X g of ethylamine is subjected to reaction with ${NaNO}_{2} / HCl$ followed by water; evolved dinitrogen gas which occupied 2.24 L volume at STP. X is ______ $\times 10^{- 1}$ g. [2024]

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Solution

Q. X g of ethylamine is subjected to reaction with NaNO2/HCl followed by water; evolved dinitrogen gas which occupied 2.24 L volume at STP. X is ______ ×10-1 g. [2024]

Ans. (45) C2H5NH2→NaNO2+HClC2H5N2+ →H2OC2H5OH+N2 Moles of nitrogen (nN2)=Volume at STP in L22.4=2.2422.4 mol=0.1 mol From reaction it is clear that moles of ethylamine (nC2H5NH2)=nN2=0.1mol. Mass of ethylamine = nC2H5NH2× molar mass of ethylamine =0.1 mol×45 g mol-1=45×10-1 g

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Q.
X g of ethylamine is subjected to reaction with ${NaNO}_{2} / HCl$ followed by water; evolved dinitrogen gas which occupied 2.24 L volume at STP. X is ______ $\times 10^{- 1}$ g. [2024]