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N2(g) + 3H2(g) → 2NH3(g) How many grams of N2 are required to produce 240.0g NH3?

just olya [345]

Answer:

$\large \boxed{\text{197.4 g}}$

Explanation:

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 28.01 17.03

N₂ + 3H₂ ⟶ 2NH₃

m/g: 240.0

(a) Moles of NH₃

$\text{Moles of NH}_{3} = \text{240.0 g NH}_{3}\times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}}= \text{14.09 mol NH}_{3}$

(b) Moles of N₂

$\text{Moles of N$_{2}$} = \text{14.09 mol NH}_{3} \times \dfrac{\text{1 mol N$_{2}$}}{\text{2 mol NH}_{3}} = \text{7.046 mol N$_{2}$}$

(c) Mass of N₂

$\text{Mass of N$_{2}$} =\text{7.046 mol N$_{2}$} \times \dfrac{\text{28.01 g N$_{2}$}}{\text{1 mol N$_{2}$}} = \textbf{197.4 g N$_{2}$}\\\\\text{The reaction requires $\large \boxed{\textbf{197.4 g}}$ of N$_{2}$}$

7 0

3 years ago

Read 2 more answers

3. Do distilled water conduct electricity? What will happen if we add sugar to it and

nikitadnepr [17]

Answer:

It cannot conduct electricity, however adding salt or sugar will make the water have impurities/other substance making it easier to conduct electricity

Explanation:

Distilled water by itself does not contain impurities, thus, it cannot conduct electricity.

When you put salt in water, the water molecules pull the sodium and chlorine ions apart so they are floating freely, increasing the conductivity.

For more information, please refer to the internet :D

Have fun studying, and goodluck!

If you are satisfied with this answer, please rate it or give brainliest.

8 0

2 years ago

Read 2 more answers

What is the name of the compound made from Sn 4+ and PO4 3-?

givi [52]

Answer:

Sn₃(PO₄)₄ - tin(IV) phosphate.

Explanation:

Hope it helps! :)

8 0

3 years ago

Na+H2O=NaOH+H2 Balancing

elena55 [62]

Answer:

Hey mate, here is your answer. Hope it helps you.

Explanation:

The balanced equation is 2Na+2H2O———-> 2NaOH +H2.

6 0

3 years ago

Hydrogen iodide, HI, is formed in an equilibrium reaction when gaseous hydrogen and iodine gas are heated together. If 20.0 g of

Kaylis [27]

Answer: D. 19.9 g hydrogen remains.

Explanation:

To calculate the moles, we use the equation:

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

a) moles of $H_2$

$\text{Number of moles}=\frac{20.0g}{2g/mol}=10.0moles$

b) moles of $I_2$

$\text{Number of moles}=\frac{20.0g}{254g/mol}=0.0787moles$

$H_2(g)+I_2(g)\rightarrow 2HI(g)$

According to stoichiometry :

1 mole of $I_2$ require 1 mole of $H_2$

Thus 0.0787 moles of $l_2$ require= $\frac{1}{1}\times 0.0787=0.0787moles$ of $H_2$

Thus $l_2$ is the limiting reagent as it limits the formation of product and $H_2$ acts as the excess reagent. (10.0-0.0787)= 9.92 moles of $H_2$ are left unreacted.

Mass of $H_2=moles\times {\text {Molar mass}}=9.92moles\times 2.01g/mol=19.9g$

Thus 19.9 g of $H_2$ remains unreacted.

5 0

3 years ago