Chemistry, 15.07.2019 11:20 luis4921

Ammonia can also be synthesized by this reaction: 3h2(g)+n2(g) → 2nh3(g) what maximum amount of ammonia in grams can be synthesized from 25.2 g of n2 and 8.42 g of h2? express your answer in grams to one decimal place.

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30.6 g First, determine the molar masses of the reactants and products. Start by looking up the atomic weights of all involved elements. Atomic weight nitrogen = 14.0067 Atomic weight hydrogen = 1.00794 Molar mass N2 = 2 * 14.0067 = 28.0134 g/mol Molar mass H2 = 2 * 1.00794 = 2.01588 g/mol Molar mass NH3 = 14.0067 + 3 * 1.00794 = 17.03052 g/mol Now determine how many moles of each reactant we have Moles nitrogen gas = 25.2 g / 28.0134 g/mol = 0.899569492 mol Moles hydrogen gas = 8.42 g / 2.01588 g/mol = 4.176835923 mol Determine the limiting reactant. The balanced equation shows that for every mole of nitrogen gas, we need 3 moles of hydrogen gas. Since we have over 4 moles of hydrogen gas and less than 1 mole of nitrogen gas, it's obvious that the limiting reactant we have is nitrogen gas. And for every mole of nitrogen gas used, we produce 2 moles of ammonia, so the expected yield is 2 * 0.899569492 mol = 1.799138983 mol. Now we just need to multiply the number of moles of ammonia by the molar mass of ammonia, so 1.799138983 mol * 17.03052 g/mol = 30.64027244 g Rounding to 1 decimal place gives 30.6 g

Answer from: maligv

$\boxed{30.62{\text{ g}}}$ of ${\text{N}}{{\text{H}}_3}$ will be produced from 25.2 g of ${{\text{N}}_{\text{2}}}$ and 8.42 g of ${{\text{H}}_{\text{2}}}$ .

Further explanation:

Limiting reagent:

It is completely consumed in a chemical reaction. It decides the amount of product formed in any chemical reaction. The amount of product depends on the amount of limiting reagent since the product formation is not possible in the absence of it.

The given balanced chemical equation for the formation of ${\text{N}}{{\text{H}}_3}$ is as follows:

$3{{\text{H}}_2}\left(g\right)+{{\text{N}}_2}\left(g\right)\to2{\text{N}}{{\text{H}}_3}\left(g\right)$

From the balanced chemical reaction, the reaction stoichiometry between ${{\text{H}}_{\text{2}}}$ and ${{\text{N}}_2}$ is as follows:

$\boxed{3{\text{ mol }}{{\text{H}}_2}:1{\text{ mol }}{{\text{N}}_2}}$

According to the reaction stoichiometry, 3 moles of ${{\text{H}}_{\text{2}}}$ and 1 mole of ${{\text{N}}_2}$ react to form 2 moles of ${\text{N}}{{\text{H}}_3}$ but the reaction mixture has 25.2 g of ${{\text{N}}_2}$ and 8.42 g of ${{\text{H}}_{\text{2}}}$ .

The formula to calculate the moles of ${{\text{H}}_{\text{2}}}$ is as follows:

${\text{Moles of }}{{\text{H}}_{\text{2}}}=\frac{{{\text{Given mass of }}{{\text{H}}_{\text{2}}}}}{{{\text{Molar mass of }}{{\text{H}}_{\text{2}}}}}$ ...... (1)

Substitute 8.42 g for the given mass of ${{\text{H}}_{\text{2}}}$ and 2.02 g/mol for the molar mass of ${{\text{H}}_{\text{2}}}$ in equation (1).

$\begin{aligned}{\text{Moles of}}{{\text{H}}_{\text{2}}}&=\left({{\text{8}}{\text{.42 g}}}\right)\left({\frac{{{\text{1 mol}}}}{{{\text{2}}{\text{.02 g}}}}}\right)\\&=4.168{\text{ mol}}\\\end{aligned}$

The formula to calculate the moles of ${{\text{N}}_2}$ is as follows:

${\text{Moles of}}{{\text{N}}_2}=\frac{{{\text{Given mass of}}{{\text{N}}_2}}}{{{\text{Molar mass of }}{{\text{N}}_2}}}$ ...... (2)

Substitute 25.2 g for the given mass of ${{\text{N}}_2}$ and 28.01 g/mol for the molar mass of ${{\text{N}}_2}$ in equation (2).

$\begin{aligned}{\text{Moles of }}{{\text{N}}_2}&=\left({{\text{25}}{\text{.2 g}}}\right)\left({\frac{{{\text{1 mol}}}}{{{\text{28}}{\text{.01 g}}}}} \right)\\&=0.899{\text{ mol}}\\\end{aligned}$

Calculate the moles of ${{\text{N}}_2}$ that react with and 4.168 moles of ${{\text{H}}_{\text{2}}}$ as follows:

$\begin{aligned}{\text{Amount of }}{{\text{N}}_2}\left( {{\text{mol}}}\right)&=\left({{\text{4}}{\text{.168 mol }}{{\text{H}}_2}}\right)\left({\frac{{1{\text{mol}}{{\text{N}}_2}}}{{{\text{3 mol }}{{\text{H}}_2}}}}\right)\\&=1.389{\text{mol}}{{\text{N}}_2}\\\end{aligned}$

The calculations concluded that 1.389 moles of ${{\text{N}}_2}$ is required for 4.168 moles of ${{\text{H}}_{\text{2}}}$ but we have only 0.899 moles of ${{\text{N}}_2}$ in the reaction mixture. Therefore, ${{\mathbf{N}}_{\mathbf{2}}}$ is present in limited quantity and is a limiting reagent.

According to the balanced chemical equation 1 mole of ${{\text{N}}_2}$ produce 2 moles of ${\text{N}}{{\text{H}}_3}$ , therefore, the moles of ${\text{N}}{{\text{H}}_3}$ produced by 1.389 moles of ${{\text{N}}_2}$ can be calculated as follows:

$\begin{aligned}{\text{Amount of N}}{{\text{H}}_3}\left({{\text{mol}}}\right)&=\left({{\text{0}}{\text{.899 mol}}{{\text{N}}_2}}\right)\left({\frac{{2{\text{mol N}}{{\text{H}}_3}}}{{1{\text{ mol }}{{\text{N}}_2}}}}\right)\\&=1.798{\text{mol N}}{{\text{H}}_{\text{3}}}\\\end{aligned}$

The formula to calculate the mass of ${\text{N}}{{\text{H}}_3}$ is as follows:

${\text{Mass of N}}{{\text{H}}_3}=\left({{\text{Moles of N}}{{\text{H}}_3}}\right)\left({{\text{Molar mass of N}}{{\text{H}}_3}}\right)$ ...... (3)

Substitute 1.798 mol for the moles of ${\text{N}}{{\text{H}}_3}$ and 17.03 g/mol for the molar mass of ${\text{N}}{{\text{H}}_3}$ in equation (3).

$\begin{aligned}{\text{Mass of N}}{{\text{H}}_3}&=\left({{\text{1}}{\text{.798 mol}}}\right)\left({\frac{{{\text{17}}{\text{.03 g}}}}{{{\text{1 mol}}}}}\right)\\&=30.619{\text{ g}}\\&\approx 30.62{\text{ g}}\\\end{aligned}$

So 30.62 g of ${\mathbf{N}}{{\mathbf{H}}_{\mathbf{3}}}$ will be produced from 25.2 g of ${{\mathbf{N}}_{\mathbf{2}}}$ and 8.42 g of ${{\mathbf{H}}_{\mathbf{2}}}$ .

Learn more:

1. Calculate the moles of chlorine in 8 moles of carbon tetrachloride: link

2. How many grams of potassium was in the fertilizer? link

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Mole concept

Keywords: limiting reagent,NH3,N2,H2,2NH3,3H2,30.62 g,2.778 mol,17.03 g/mol,2.02 g/mol,1.389 mol,25.2 g,8.42 g,4.168 mol,0.899 mol,1 mole,2 moles,3 moles.

Answer from: Quest

by cartels in other countries, shipped by air, mail, or land. i cant really see your answer choices : /

Answer from: Quest

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally

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Ammonia can also be synthesized by this reaction: 3h2(g)+n2(g) → 2nh3(g) what maximum amount of amm...

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