Chemistry, 12.02.2020 03:48 srtdklug2005

If 65.0 mL of 0.150 M HCl ( aq ) is needed to neutralize all the NH 3 ( g ) from a 2.25 g sample of organic material, calculate the mass percentage of nitrogen in the sample.

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Answer from: kikirogers6530

The mass percentage of nitrogen in the sample is 6.04 %

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of HCl solution = 0.150 M

Volume of solution = 65.0 mL = 0.065 L (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$0.150M=\frac{\text{Moles of HCl}}{0.065L}\\\\\text{Moles of HCl}=(0.150mol/L\times 0.065L)=9.75\times 10^{-3}mol$

The chemical equation for the reaction of HCl and ammonia follows:

$HCl+NH_3\rightarrow NH_4Cl$

By Stoichiometry of the reaction:

1 mole of HCl reacts with 1 mole of ammonia

So, $9.75\times 10^{-3}mol$ of HCl will react with = $\frac{1}{1}\times 9.75\times 10^{-3}mo=9.75\times 10^{-3}mol$ of ammonia

1 mole of ammonia contains 1 mole of nitrogen and 3 moles of hydrogen element

Moles of nitrogen in ammonia = $9.75\times 10^{-3}mol$

To calculate the mass for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of nitrogen = $9.75\times 10^{-3}mol$

Molar mass of nitrogen = 14 g/mol

Putting values in above equation, we get:

$9.75\times 10^{-3}mol=\frac{\text{Mass of nitrogen}}{14g/mol}\\\\\text{Mass of nitrogen}=(9.75\times 10^{-3}mol\times 14g/mol)=0.136g$

To calculate the mass percentage of nitrogen in the sample, we use the equation:

$\text{Mass percent of nitrogen}=\frac{\text{Mass of nitrogen}}{\text{Mass of sample}}\times 100$

Mass of sample= 2.25 g

Mass of nitrogen = 0.136 g

Putting values in above equation, we get:

$\text{Mass percent of nitrogen}=\frac{0.136g}{2.25g}\times 100=6.04\%$

Hence, the mass percentage of nitrogen in the sample is 6.04 %

Answer from: Quest

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

Answer from: Quest

the answer is na (sodium)

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If 65.0 mL of 0.150 M HCl ( aq ) is needed to neutralize all the NH 3 ( g ) from a 2.25 g sample of...

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