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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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Read 2 more answers

Magnesium burns in air with a dazzling brilliance to produce magnesium oxide: 2 Mg(s) + O2(g) → 2 MgO(s) How many moles of O2 ar

TEA [102]

Answer:

1.05 mols of O2 gas

Explanation:

For this type of problem, it's important to understand what the balanced Chemical equation tells us:

<h3><u>Balanced Chemical equations</u></h3>

Let's look at the balanced chemical equation:

$2Mg(s)+O_2(g)-- > 2MgO(s)$

This equation two sides of the reaction arrow. On the left are the reactants (things you start with, and that react during the chemical reaction), and on the right side are products (things that are produced during the chemical reaction that you end with).

The numbers in front of each compound tell how many of molecules are involved for a full reaction without anything left over. A "mol" is a large quantity ( $6.022*10^{23}$ ), in this case, of molecules , since it's unlikely you're only taking a single molecule of each substance (it would be so tiny, you wouldn't even know you were doing the reaction).

So,for every 2 moles of Magnesium used, we'll also need 1 mole of Oxygen, and it will produce 2 moles of Magnesium Oxide.

In a way, during the reaction it's almost like 2 moles of Magnesium is equal to 1 mole of Oxygen and is equal to 2 moles of Magnesium Oxide:

$2 mol Mg(s)=1mol O_2(g)=2molMgO(s)$

From here, we can build some unit ratios, to convert between the known quantity of moles we have, and find the unknown quantity of moles that are requested.

<h3><u>Finding the right unit ratio</u></h3>

We know that we are looking at 2.10 mol of Magnesium, so we want a unit ratio with moles of Mg on the bottom. <u>We want to find moles of O2</u>, <u>so we want a unit ratio of moles of O2 on top</u>.

The unit ratio we want is the middle part of the equation, divided by the left part of the equation.

$2 mol Mg(s)=1mol O_2(g)\\\frac{2 mol Mg(s)}{2 mol Mg(s)}=\frac{1mol O_2(g)}{2 mol Mg(s)}\\1=\frac{1mol O_2(g)}{2 mol Mg(s)}$

Since this quantity is 1, it is a unit ratio and can be multiplied to other things to change their units (for this problem).

<h3><u>Finding the answer</u></h3>

Starting with what we know, and multiplying by our unit ratio:

$2.10 mol Mg(s)*\frac{1mol O_2(g)}{2 mol Mg(s)}=1.05mol O_2(g)$

Notice that the units from the first quantity cancel with the units on the bottom of the fraction, leaving only the unit on top of the fraction ... the exact units we wanted!

So, 1.05mols of O2 would be consumed during the reaction if exactly 2.10moles of Magnesium are burned.

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2 years ago