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Rudik [331]
3 years ago
7

Which equations represent precipitation reactions?

Chemistry
1 answer:
viktelen [127]3 years ago
3 0

In a precipitation reaction, one of the product formed is an insoluble precipitate.

The given reactions can be written with phases included as:

A. Na_{2}S(aq)+FeBr_{2}(aq)-->2NaBr(aq)+FeS(s): In this reaction one of the product, FeS is insoluble. Therefore, this is a precipitation reaction.

B. MgSO_{4}(aq)+CaCl_{2}(aq)-->MgCl_{2}(aq)+CaSO_{4}(s): In this reaction, the product CaSO_{4}is a solid(insoluble). So, this is a precipitation reaction too.

C.LiOH(aq)+NH_{4}I(aq)-->LiI(aq)+NH_{4}OH(aq): In this reaction, both the products are soluble. So this is not a precipitation reaction.

D.2NaCl(aq)+K_{2}S(aq)-->Na_{2}S(aq)+2KCl(aq): In this reaction, both the products are soluble. So this is not a precipitation reaction.

E. AgNO_{3}(aq)+NaCl(aq)-->AgCl(s)+NaNO_{3}(aq): In this reaction, the product AgCl is a precipitate. So, it is a precipitation reaction.


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1. To solve this question we need to find the volume of the rectangle. With the volume and density we can find the mass of the solid:

Volume = 7.45cm*4.78cm*5.25cm

Volume = 187cm³

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187cm³ * (2.702g/cm³) = 505g is the mass of the aluminium

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1) 65.0

2) 16.434 L = 16434 mL.

Explanation:

<em>2NaN₃ → 2Na + 3N₂,</em>

  • It is clear from the balanced equation that 2.0 moles of NaN₃ are decomposed to 2.0 moles of Na and 3.0 moles of N₂.

<em>Q1: How many grams of NaN₃ are needed to make 23.6L of N₂?​ </em>

Density of N₂ = 0.92 g/L which means that every 1.0 L of N₂ contains 0.92 g of N₂.

  • Now, we can get the mass of N₂ in 23.6 L N₂ using cross multiplication:

1.0 L of N₂ contains → 0.92 g of N₂.

23.6 L of N₂ contains → ??? g of N₂.

∴ The mass of N₂ in 23.6 L of N₂ = (23.6 L)(0.92 g)/(1.0 L) = 21.712 g.

  • We can get the no. of moles of 23.6 L of N₂ (21.712 g) using the relation:

n = mass/molar mass = (21.712 g)/(28.0 g/mol) = 0.775 mol.

  • We can get the no. of moles of NaN₃ needed to produce 0.775 mol of N₂:

<em><u>using cross multiplication:</u></em>

2.0 moles of NaN₃ produce → 3.0 moles of N₂, from the balanced equation.

??? mol of NaN₃ produce → 0.775 moles of N₂.

∴ The no. of moles of NaN₃ needed = (2.0 mol)(0.775 mol)/(3.0 mol) = 0.517 mol.

  • Finally, we can get the grams of NaN₃ needed:

<em>mass = no. of moles x molar mass</em> = (0.517 mol)(65.0 g/mol) =<em> 33.6 g.</em>

<em />

<em>Q2: How many mL of N₂ result if 8.3 g Na are also produced?</em>

  • We need to get the no. of moles of 8.3 g Na using the relation:

n = mass/atomic mass = (8.3 g)/(22.98 g/mol) = 0.36 mol.

  • We can get the no. of moles of N₂ produced with 0.36 mol of Na:

<em><u>using cross multiplication:</u></em>

2.0 moles of Na produced with → 3.0 moles of N₂, from the balanced equation.

0.36 moles of Na produced with → ??? moles of N₂.

∴ The no. of moles of N₂ needed = (3.0 mol)(0.36 mol)/(2.0 mol) = 0.54 mol.

  • We can get the mass of 0.54 mol of N₂:

mass = no. of moles  x molar mass = (0.54 mol)(28.0 g/mol) = 15.12 g.

  • Now, we can get the mL of 15.12 g of N₂:

<em><u>using cross multiplication:</u></em>

1.0 L of N₂ contains → 0.92 g of N₂, from density of N₂ = 0.92 g/L.

??? L of N₂ contains → 15.12 g of N₂.

<em>∴ The volume of N₂ result </em>= (1.0 L)(15.12 g)/(0.92 g) = <em>16.434 L = 16434 mL.</em>

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