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3 years ago

How many moles of Ammonium Sulfate can be made from 30.0mol of NH3? 2NH3 + H2SO4 --> (NH4)2SO4

Chemistry

1 answer:

BartSMP [9]3 years ago

7 0

Answer: 15.0 moles of $(NH_4)_2SO_4$ are formed from 30.0 mol of $NH_3$

Explanation:

The balanced chemical reaction is :

$2NH_3+H_2SO_4\rightarrow (NH_4)_2SO_4$

According to stoichiometry :

2 moles of $NH_3$ give = 1 mole of $(NH_4)_2SO_4$

Thus 30.0 moles of $NH_3$ will give = $\frac{1}{2}\times 30.0=15.0moles$ of $(NH_4)_2SO_4$

Thus 15.0 moles of $(NH_4)_2SO_4$ are formed from 30.0 mol of $NH_3$

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How much energy is released by the decay of 3 grams of 230Th in the following reaction 230 Th - 226Ra + "He (230 Th = 229.9837 g

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<u>Answer:</u> The energy released for the decay of 3 grams of 230-Thorium is $2.728\times 10^{-15}J$

<u>Explanation:</u>

First we have to calculate the mass defect $(\Delta m)$ .

The equation for the alpha decay of thorium nucleus follows:

$_{90}^{230}\textrm{Th}\rightarrow _{88}^{226}\textrm{Ra}+_2^{4}\textrm{He}$

To calculate the mass defect, we use the equation:

Mass defect = Sum of mass of product - Sum of mass of reactant

$\Delta m=(m_{Ra}+m_{He})-(m_{Th})$

$\Delta m=(225.9771+4.008)-(229.9837)=1.4\times 10^{-3}amu=2.324\times 10^{-30}kg$

(Conversion factor: $1amu=1.66\times 10^{-27}kg$ )

To calculate the energy released, we use Einstein equation, which is:

$E=\Delta mc^2$

$E=(2.324\times 10^{-30}kg)\times (3\times 10^8m/s)^2$

$E=2.0916\times 10^{-13}J$

The energy released for 230 grams of decay of thorium is $2.0916\times 10^{-13}J$

We need to calculate the energy released for the decay of 3 grams of thorium. By applying unitary method, we get:

As, 230 grams of Th release energy of = $2.0916\times 10^{-13}J$

Then, 3 grams of Th will release energy of = $\frac{2.0916\times 10^{-13}}{230}\times 3=2.728\times 10^{-15}J$

Hence, the energy released for the decay of 3 grams of 230-Thorium is $2.728\times 10^{-15}J$

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