What is the mass of 2 moles of 238 u?

Chemistry

2 answers:

Ad libitum [116K]2 years ago

5 0

<span>use this formula mass = moles x molar mass</span>

Lana71 [14]2 years ago

3 0

Answer:

Mass of 2 moles of $^{238}\textrm{U}$ is 476.1 g

Explanation:

We know that number of moles of an atom is ratio of mass of the atom to it's atomic weight.
Atomic weight of $^{238}\textrm{U}$ is 238.05 g/mol
So, mass of 1 mol of $^{238}\textrm{U}$ atom is 238.05 g
Hence mass of 2 moles of $^{238}\textrm{U}$ = $(2mol\times 238.05g/mol)=476.1 g$

Hence mass of 2 moles of $^{238}\textrm{U}$ is 476.1 g

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If you feed 100 kg of N2 gas and 100 kg of H2 gas into a

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Answer : The mass of ammonia produced can be, 121.429 k

Solution : Given,

Mass of $N_2$ = 100 kg = 100000 g

Mass of $H_2$ = 100 kg = 100000 g

Molar mass of $N_2$ = 28 g/mole

Molar mass of $H_2$ = 2 g/mole

Molar mass of $NH_3$ = 17 g/mole

First we have to calculate the moles of $N_2$ and $H_2$ .

$\text{ Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molar mass of }N_2}=\frac{100000g}{28g/mole}=3571.43moles$

$\text{ Moles of }H_2=\frac{\text{ Mass of }H_2}{\text{ Molar mass of }H_2}=\frac{100000g}{2g/mole}=50000moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$N_2+3H_2\rightarrow 2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $N_2$ react with 3 mole of $H_2$

So, 3571.43 moles of $N_2$ react with $3571.43\times 3=10714.29$ moles of $H_2$

From this we conclude that, $H_2$ is an excess reagent because the given moles are greater than the required moles and $N_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NH_3$

From the reaction, we conclude that

As, 1 mole of $N_2$ react to give 2 mole of $NH_3$

So, 3571.43 moles of $N_2$ react to give $3571.43\times 2=7142.86$ moles of $NH_3$