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2 years ago

What is the percent composition by mass of nitrogen in the compound N2H4 (gram-formula mass = 32 g/mol)?

Chemistry

2 answers:

Karolina [17]2 years ago

8 0

Answer:88%

Explanation:

Alex2 years ago

3 0

Answer:

$\large \boxed{93\, \% }$

Explanation:

1. Calculate the molar mass of N₂H₄

2N = 2 × 14 = 28

4H = 2 × 1 = <u> 4</u>

Tot. = 32

2. Calculate the mass percent of N

$\text{\% of element} = \dfrac{\text{mass of element}}{\text{mass of compound}} \times 100 \, \% = \dfrac{\text{28}}{\text{32}} \times 100 \, \% =\mathbf{88 \, \%}\\\\\text{The percentage of N in N$_{2}$H$_{4}$ is $\large \boxed{\mathbf{88\, \% }}$}}$

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Explanation:

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Volume of water = 35 L = 35 × 10³ mL

initial temperature of water = 25.0 ° C

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$q_{boiling} = 10.9725 \times 10^6 \ J$

Also; Assume that the fuel has an average formula of C7 H16 and 15% of the heat generated from combustion goes to heat the water;

thus the heat of combustion can be determined via the expression

$q_{combustion} =- \dfrac{q_{boiling}}{0.15}$

$q_{combustion} =- \dfrac{10.9725 \times 10^6 J}{0.15}$

$q_{combustion} = -7.315 \times 10^{7} \ J$

$q_{combustion} = -7.315 \times 10^{4} \ kJ$

For heptane; the equation for its combustion reaction can be written as:

$C_7H_{16} + 11O_{2(g)} -----> 7CO_{2(g)}+ 8H_2O_{(g)}$

The standard enthalpies of the products and the reactants are:

$\Delta H _f \ CO_{2(g)} = -393.5 kJ/mol$

$\Delta H _f \ H_2O_{(g)} = -242 kJ/mol$

$\Delta H _f \ C_7H_{16 }_{(g)} = -224.4 kJ/mol$

$\Delta H _f \ O_{2{(g)}} = 0 kJ/mol$

Therefore; the standard enthalpy for this combustion reaction is:

$\Delta H ^0= \sum n_p\Delta H^0_{f(products)}- \sum n_r\Delta H^0_{f(reactants)}$

$\Delta H^0 =( 7 \ mol ( -393.5 \ kJ/mol) + 8 \ mol (-242 \ kJ/mol) -1 \ mol( -224.4 \ kJ/mol) - 11 \ mol (0 \ kJ/mol))$

$\Delta H^0 = (-2754.5 \ \ kJ - 1936 \ \ kJ+224.4 \ \ kJ+0 \ \ kJ)$

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$n_{fuel} = \dfrac{q}{\Delta \ H^0}$

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$n_{fuel} = 16.38 \ mol \ of \ C_7 H_{16$

Since number of moles = mass/molar mass

The mass of the fuel is:

$m_{fuel } = 16.38 mol \times 100.198 \ g/mol}$

$m_{fuel } = 1.641 \times 10^{3} \ g$

Given that the density of the fuel is = 0.78 g/mL

and we know that :

density = mass/volume

therefore making volume the subject of the formula in order to determine the volume of the fuel ; we have

volume of the fuel = mass of the fuel / density of the fuel

volume of the fuel = $\dfrac{1.641 \times 10^3 \ g }{0.78 g/mL} \times \dfrac{L}{10^3 \ mL}$

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