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3 years ago

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When 4.15 grams of silver nitrate is reacted with 1.11 grams of iron(III) chloride, which best represents the amount of silver c

hloride produced?

1 answer:

PolarNik [594]3 years ago

3 0

Answer:

The mass of silver chloride produced = 2.202 g

Explanation:

Equation of the reaction is given below

3AgNO₃(aq) + FeCl₃(aq) ----> 3AgCl(s) + Fe(NO₃)₃(aq)

molar mass of AgNO₃ = 170 g/mol

molar mass of FeCl₃ = 233.5 g/mol

molar mass of AgCl = 143.5 g/mol

3 moles of silver nitrate reacts with 1 mole of iron (iii) chloride to give 3 moles of silver nitrate

4.15 grams of AgNO₃ = 4.15/170 = 0.0244 moles of AgNO₃

1.11 grams of FeCl₃ = 1.11/233.5 = 0.0047 moles of FeCl₃

mole ratio of AgNO₃ to FeCl₃ = 0.0244/0.0047 = 5 : 1

therefore, FeCl₃ is the limiting reactant

0.0047 moles of FeCl₃ reacting will produce 0.0047 * 3 moles of AgCl = 0.0141 moles of AgCl

0.0141 moles of AgCl = 0.0141 * 143.5 g of AgCl = 2.02 g of AgCl =

Therefore mass of silver chloride produced = 2.202 g

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A chemist wants to calculate the amount of heat that is absorbed by a sample of copper as it is melted. Which constant should sh

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The constant used for the absorption of heat by the sample in melting is $+\Delta H_{fus}$ . Thus, option A is correct.

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(a) What is the total volume (in L) of gaseous products, measured at 350°C and 735 torr, when an automobile engine burns 100. g

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Answer:

Part A

The volume of the gaseous product is $V = 787L$

Part B

The volume of the the engine’s gaseous exhaust is $V_e = 2178 \ L$

Explanation:

Part A

From the question we are told that

The temperature is $T = 350^oC = 350 +273 =623K$

The pressure is $P = 735 \ torr = \frac{735}{760} = 0.967\ atm$

The of $C_8 H_{18} = 100.0g$

The chemical equation for this combustion is

$2 C_8 H_{18}_{(l)} + 25O_2_{(l)} ----> 16CO_2_{(g)} + 18 H_2 O_{(g)}$

The number of moles of $C_8 H_{18}$ that reacted is mathematically represented as

$n = \frac{mass \ of \ C_8H_{18} }{Molar \ mass \ of C_8H_{18} }$

The molar mass of $C_8 H_{18}$ is constant value which is

$M = 114.23 \ g/mole$

So $n = \frac{100 }{114.23} }$

$n = 0.8754 \ moles$

The gaseous product in the reaction is $CO_2_{(g)}$ and water vapour

Now from the reaction

2 moles of $C_8 H_{18}$ will react with 25 moles of $O_2$ to give (16 + 18) moles of $CO_2_{(g)}$ and $H_2 O_{(g)}$

So

1 mole of $C_8 H_{18}$ will react with 12.5 moles of $O_2$ to give 17 moles of $CO_2_{(g)}$ and $H_2 O_{(g)}$

This implies that

0.8754 moles of $C_8 H_{18}$ will react with (12.5 * 0.8754 ) moles of $O_2$ to give (17 * 0.8754) of $CO_2_{(g)}$ and $H_2 O_{(g)}$

So the no of moles of gaseous product is

$N_g = 17 * 0.8754$

$N_g = 14.88 \ moles$

From the ideal gas law

$PV = N_gRT$

making V the subject

$V = \frac{N_gRT}{P}$

Where R is the gas constant with a value $R = 0.08206 \ L\cdot atm /K \cdot mole$

Substituting values

$V = \frac{14.88* 0.08206 *623}{0.967}$

$V = 787L$

Part B

From the reaction the number of moles of oxygen that reacted is

$N_o = 0.8754 * 12.5$

$N_o = 10.94 \ moles$

The volume is

$V_o = \frac{10.94 * 0.08206 *623}{0.967}$

$V_o = 579 \ L$

No this volume is the 21% oxygen that reacted the 79% of air that did not react are the engine gaseous exhaust and this can be mathematically evaluated as

$V_e = V_o * \frac{0.79}{0.21}$

Substituting values

$V_e = 579 * \frac{0.79}{0.21}$

$V_e = 2178 \ L$

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3 years ago