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2 years ago

1. Given the following equation:

1 answer:

5 0

The percent yield obtained from the stoichiometry of the reaction when the moles of the reactants and products are known.

<h3>What is stoichiometry?</h3>

Stochiometry gives the relationship between the mass and mole or moles and volume in a reaction.

Q1)

K2PtCl4 + 2NH3 ----> Pt(NH3)2Cl2 +2KCl

Number of moles of NH3 = 34.5 g/17 g/mol = 2.03 moles

2 moles of NH3 yields 2 moles of KCl

2.03 moles of NH3 yields 2.03 moles of KCl

Theoretical yield of KCl = 2.03 moles * 75 g/mol = 154 g

Again,

2moles of NH3 yields 1 mole of Pt(NH3)2Cl2

2.03 moles of NH3 yields 2.03 moles * 1/2 = 1.015 moles

Mass of Pt(NH3)2Cl2 = 1.015 moles * 301 g/mol = 306 g

% yield = 76.4 g/306 g * 100/1 = 25%

Q2)

H3PO4 + 3 KOH ------> K3PO4 + 3 H2O

Number of moles of H3PO4 = 49.0 g/98 g/mol = 0.5 moles

If mole of H3PO4 yield 1 mole of K3PO4

0.5 moles of H3PO4 yield 0.5 moles of K3PO4

Mass of K3PO4 = 212 g/mol * 0.5 moles = 106 g

Percent yield = 49.0 g/106 g * 100 = 46%

Q3)

Al2(SO3)3 + 6 NaOH ------> 3 Na2SO3 + 2 Al(OH)3

Number of moles of Al2(SO3)3 = 389.4 g/294 g/mol = 1.32 moles

If 1 mole of Al2(SO3)3 yields 3 moles of Na2SO3

1.32 moles of Al2(SO3)3 yields 1.32 moles * 3 moles/1 mole = 3.96 moles

Mass of Na2SO3 = 3.96 moles * 126 g/mol = 498.96 g

Percent yield = 212.4 g/498.96 g * 100/1 = 43%

Q4)

Al(OH)3(s) + 3 HCl(aq) -------> AlCl3(aq) + 3 H2O(l)

Number of moles of Al(OH)3 = 50.3 g/78 g/mol = 0.64 moles

If 1 mole of Al(OH)3 yields 1 mole of AlCl3

0.64 moles of Al(OH)3 yields 0.64 moles of AlCl3

Mass of AlCl3 = 133 g/mol * 0.64 moles = 85 g

Percent yield = 39.5 g/85 g * 100 = 46%

Q5) K2CO3 + 2HCl --------> H2O + CO2 + 2KCl

Number of moles of K2CO3 = 34.5 g/138 g/mol = 0.25 moles

1 mole of K2CO3 produces 2 moles of KCl

0.25 moles of K2CO3 produces 0.25 moles * 2 moles/1 mole = 0.5 moles

Mass of KCl = 0.5 moles * 75 g/mol = 37.5 g

If 1 mole of K2CO3 yields 1 mole of H2O

0.25 moles of K2CO3 yields 0.25 moles of H2O

Mass of H2O = 0.25 moles * 18 g/mol = 4.5 g

Percent yield = 3.4 g/4.5 g *100 = 76%

Learn more about percent yield: brainly.com/question/12704041?

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3 years ago

A solution is prepared by mixing 50.0 mL toluene (C6H5CH3 d=0.867 g/mL) with 125 mL Benzene (C6H6 d=0.874 g/mL). Assuming that t

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Answer:

mass % = 28.4%

mole fraction = 0.252

molality = 3.08 molal

molarity = 2.69 M

Explanation:

Step 1: Data given

Volume of toluene = 50.0 mL = 0.05 L

Density toluene = 0.867 g/mL

Molar mass toluene = 92.14 g/mol

Volume of benzene = 125 mL = 0.125 L

Density benzene = 0.874 g/mL

Molar mass benzene = 78.11 g/mol

Step 2: Calculate masses

Mass = density * volume

Mass toluene = 50.0 mL * 0.867 g/mL = 43.35 g

Mass benzene = 125 mL * 0.874 g/mL = 109.25 g

Step 3: Calculate number of moles

Moles = mass / molar mass

Moles toluene = 43.35 grams /92.14 g/mol = 0.470 5 moles

Moles benzene = 109.25 grams / 78.11 g/mol = 1.399 moles

Step 4: Calculate molarity of toluene

Molarity = moles / volume

Molarity toluene = 0.4705 moles / 0.175 L = 2.69 M

Step 5: Calculate mass % of toluene

Mass % = (43.35 grams / (43.35 + 109.25) )*100 % = 28.4 %

Step 6: Calculate mole fraction of toluene

Mole fraction toluene = Moles toluene / total number of moles

Mole fraction toluene = 0.4705 / (0.4705 + 1.399) = 0.252

Step 7: Molality of toluene

Molality = number of moles / mass

Molality of toluene = 0.4705 moles / (0.04335 + 0.10925)

Molality of toluene = 3.08 molal

4 0

3 years ago

An excess of sodium carbonate, Na, CO3, in solution is added to a solution containing 17.87 g CaCl2. After performing the

Brrunno [24]

Answer:

Approximately $81.84\%$ .

Explanation:

Balanced equation for this reaction:

${\rm Na_{2}CO_{3}}\, (aq) + {\rm CaCl_{2}} \, (aq) \to 2\; {\rm NaCl}\, (aq) + {\rm CaCO_{3}}\, (s)$ .

Look up the relative atomic mass of elements in the limiting reactant, $\rm CaCl_{2}$ , as well as those in the product of interest, $\rm CaCO_{3}$ :

$\rm Ca$ : $40.078$ .
$\rm Cl$ : $35.45$ .
$\rm C$ : $12.011$ .
$\rm O$ : $15.999$ .

Calculate the formula mass for both the limiting reactant and the product of interest:

$\begin{aligned}& M({\rm CaCl_{2}}) \\ &= (40.078 + 2 \times 35.45)\; {\rm g \cdot mol^{-1}} \\ &= 110.978\; \rm g \cdot mol^{-1}\end{aligned}$ .

$\begin{aligned}& M({\rm CaCO_{3}}) \\ &= (40.078 + 12.011 + 3 \times 15.999)\; {\rm g \cdot mol^{-1}} \\ &= 100.086\; \rm g \cdot mol^{-1}\end{aligned}$ .

Calculate the quantity of the limiting reactant ( $\rm CaCl_{2}$ ) available to this reaction:

$\begin{aligned}n({\rm CaCl_{2}) &= \frac{m({\rm {CaCl_{2}})}}{M({\rm CaCl_{2}})} \\ &= \frac{17.87\; \rm g}{110.978\; \rm g \cdot mol^{-1}} \\ &\approx 0.161023\; \rm mol \end{aligned}$ .

Refer to the balanced equation for this reaction. The coefficients of the limiting reactant ( $\rm CaCl_{2}$ ) and the product ( ${\rm CaCO_{3}}$ ) are both $1$ . Thus:

$\displaystyle \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} = 1$ .

In other words, for every $1\; \rm mol$ of $\rm CaCl_{2}$ formula units that are consumed, $1\; \rm mol\!$ of $\rm CaCO_{3}$ formula units would (in theory) be produced. Thus, calculate the theoretical yield of $\rm CaCO_{3}\!$ in this experiment:

$\begin{aligned} & n(\text{${\rm CaCO_{3}}$, theoretical}) \\ =\; & n({\rm CaCl_{2}}) \cdot \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} \\ \approx \; & 0.161023\; {\rm mol} \times 1 \\ =\; & 0.161023\; \rm mol\end{aligned}$ .

Calculate the theoretical yield of this experiment in terms of the mass of $\rm CaCO_{3}$ expected to be produced:

$\begin{aligned} & m(\text{${\rm CaCO_{3}}$, theoretical}) \\ = \; & n(\text{${\rm CaCO_{3}}$, theoretical}) \cdot M(({\rm CaCO_{3}}) \\ \approx \; & 0.161023\; {\rm mol} \times 100.086\; {\rm g \cdot mol^{-1}} \\ \approx \; & 16.1161\; \rm g \end{aligned}$ .

Given that the actual yield in this question (in terms of the mass of $\rm CaCO_{3}$ ) is $13.19\; \rm g$ , calculate the percentage yield of this experiment:

$\begin{aligned} & \text{percentage yield} \\ =\; & \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \\ \approx \; & \frac{13.19\; {\rm g}}{16.1161\; {\rm g}} \times 100\% \\ \approx \; & 81.84\%\end{aligned}$ .

6 0

2 years ago