The following reaction is used to extract gold from pre-treated gold ore: \(2 \mathrm{Au}(\mathrm{CN})_{2}^{-}(a q)+\mathrm{Zn}(s) \longrightarrow 2 \mathrm{Au}(s)+\mathrm{Zn}(\mathrm{CN})_{4}^{-}(a q)\) (a) How many grams of \(Z n\) are needed to react with 0.11 mol of \(\mathrm{Au}(\mathrm{CN})_{2}^{-} ?\) (b) How many grams of Au can form from \(0.11 \mathrm{~mol}\) of \(\mathrm{Au}(\mathrm{CN})_{2}^{-} ?\) (c) How many grams of \(\mathrm{Au}(\mathrm{CN})_{2}^{-}\) are required for the reaction of \(0.11 \mathrm{~mol}\) of \(\mathrm{Zn}\) ?

Short Answer

Expert verified
For (a) 3.57 g of Zn is needed. For (b) 21.69 g of Au can form. For (c) 49.70 g of \(\mathrm{Au}(\mathrm{CN})_{2}^{-}\) are required.

Step by step solution

01

Calculate the moles of Zn required

From the balanced equation, 1 mole of Zn reacts with 2 moles of \(\mathrm{Au}(\mathrm{CN})_{2}^{-}\). Therefore, the moles of Zn required for 0.11 moles of \(\mathrm{Au}(\mathrm{CN})_{2}^{-}\) are 0.11 / 2.
02

Convert the moles of Zn to grams

Using the molar mass of \(\mathrm{Zn}\) which is approximately 65.38 g/mol, the mass of \(\mathrm{Zn}\) needed can be calculated as \(0.11 \text{ moles} / 2 \times 65.38 \text{ g/mol}\).
03

Calculate the grams of Au produced

The balanced equation shows that 2 moles of \(\mathrm{Au}(\mathrm{CN})_{2}^{-}\) produce 2 moles of Au. The mass of Au formed from 0.11 moles of \(\mathrm{Au}(\mathrm{CN})_{2}^{-}\) can be calculated using the molar mass of Au which is approximately 196.97 g/mol. The mass of Au will be \(0.11 \times 196.97 \text{ g/mol}\).
04

Determine grams of \(\mathrm{Au}(\mathrm{CN})_{2}^{-}\) required for the reaction with \(\mathrm{Zn}\)

The molar mass of \(\mathrm{Au}(\mathrm{CN})_{2}^{-}\) is needed to perform this calculation. It is the sum of the atomic masses of 2 gold atoms, 4 carbon atoms, and 2 nitrogen atoms. Use this molar mass to convert 0.11 moles of \(\mathrm{Zn}\) to grams of \(\mathrm{Au}(\mathrm{CN})_{2}^{-}\) required for reaction, taking into account the 2:1 molar ratio between \(\mathrm{Au}(\mathrm{CN})_{2}^{-}\) and \(\mathrm{Zn}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mole to Gram Conversion
Understanding mole to gram conversion is vital for students tackling stoichiometry problems. In chemistry, the mole is a unit used to measure the amount of a substance. However, when working in a laboratory or calculating the outcomes of reactions, we often need to know the mass of a substance in grams.

To convert moles to grams, you need to know the substance's molar mass. The molar mass of an element is the mass of one mole of that substance and is usually measured in grams per mole (g/mol). It can be found on the periodic table as the atomic weight of the element. For compounds, the molar mass is the sum of the molar masses of its component atoms.

Here's a simple formula for converting moles to grams:
\[ \text{Mass in grams} = \text{moles} \times \text{molar mass} \]

For example, if you have 0.5 moles of a substance with a molar mass of 44 g/mol, the mass in grams would be:
\[ 0.5 \text{ moles} \times 44 \text{ g/mol} = 22 \text{ grams} \]

This conversion is crucial when predicting how much of a reactant is needed or the quantity of product formed in a chemical reaction.
Chemical Equations
Chemical equations are symbolic representations of chemical reactions. They depict the reactants, products, and their quantities. A balanced chemical equation ensures that the law of conservation of mass is obeyed; the number of atoms of each element is the same on both sides of the equation.

A stoichiometrically balanced equation reflects the molar ratios between the reactants and products, which are vital for quantitative predictions in chemistry. For example, in the equation
\[2 \mathrm{H}_2 + \mathrm{O}_2 \rightarrow 2 \mathrm{H}_2\mathrm{O}\]

it shows that two moles of hydrogen gas react with one mole of oxygen gas to form two moles of water.

Understanding these ratios allows chemists to calculate how much of each reactant is needed to produce a desired amount of product, or how much product can be made from given amounts of reactants. This concept is not only fundamental in academic exercises but also in industrial applications where efficiency and cost-effectiveness are key.
Molar Mass
Molar mass is a property that links the mass of a substance to its amount in moles and is expressed in grams per mole (g/mol). It's a necessary concept when dealing with mole to gram conversions and balancing chemical equations.

To determine the molar mass of a single element, look at the atomic mass listed on the periodic table. For instance, carbon has an atomic mass of approximately 12.01, so its molar mass is 12.01 g/mol.

In the case of a compound, you'll need to sum the molar masses of all the atoms present in the molecule. Let's take water (\(\mathrm{H}_2\mathrm{O}\)) as an example:
  • The molar mass of hydrogen is about 1.01 g/mol, and since there are two hydrogen atoms, their combined mass is 2.02 g/mol.
  • Oxygen's molar mass is about 16.00 g/mol.
  • So, the molar mass of water is \(2 \times 1.01 + 16.00 = 18.02\) g/mol.

With the molar mass, you can now easily convert between moles and grams to understand the quantities involved in chemical reactions, which is pivotal in both academic problems and real-world chemical processes.

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Most popular questions from this chapter

A compound of \(\mathrm{Ca}, \mathrm{C}, \mathrm{N},\) and \(\mathrm{S}\) was subjected to quantitative analysis and formula mass determination, and the following data were obtained. A \(0.250 \mathrm{~g}\) sample was mixed with \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) to convert all of the Ca to \(0.160 \mathrm{~g}\) of \(\mathrm{CaCO}_{3}\). A \(0.115 \mathrm{~g}\) sample of the compound was carried through a series of reactions until all of its \(S\) was changed to \(0.344 \mathrm{~g}\) of \(\mathrm{BaSO}_{4}\). A \(0.712 \mathrm{~g}\) sample was processed to liberate all of its \(\mathrm{N}\) as \(\mathrm{NH}_{3}\), and \(0.155 \mathrm{~g}\) of \(\mathrm{NH}_{3}\) was obtained. The formula mass was found to be 156 . Determine the empirical and molecular formulas of this compound.

It was found that \(2.35 \mathrm{~g}\) of a compound of phosphorus and chlorine contained \(0.539 \mathrm{~g}\) of phosphorus. What are the percentages by mass of phosphorus and chlorine in this compound?

Which has a higher percentage of nitrogen: carbamazepine, an anticonvulsant, \(\left(\mathrm{C}_{15} \mathrm{H}_{12} \mathrm{~N}_{2} \mathrm{O}\right)\) or carbetapentane, a cough suppressant, \(\left(\mathrm{C}_{20} \mathrm{H}_{31} \mathrm{NO}_{3}\right)\) ?

How many grams of \(\mathrm{C}\) are combined with \(4.25 \times 10^{23}\) atoms of \(\mathrm{H}\) in the compound \(\mathrm{C}_{5} \mathrm{H}_{12} ?\)

A balanced chemical equation contains the term \(“ 2 \mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O} .\) " How many atoms of each element does this represent in the molecular view? How many moles of each element does this represent on the laboratory scale?

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