Question 1: How many grams of ammonium nitrate are there in 4.524 mols of the compound?

Question

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Question 2: How many molecules are there in 2.15 mols of ammonium nitrate?
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Question 3: When a sample of solid silver bromide decomposes over time it will form solid silver and bromine gas. This decomposition does reach completion and all reactant transitions to products.

A. Write the full reaction with the correct physical states and balance the equation with all required work/steps or explanation, if needed.

B. How many grams of bromine gas will be produced when 25.54 grams of silver bromide solid decomposes fully? To earn points you must show the correct stoichiometric set up for calculation. See class notes for needed examples.

All questions must be 2-5 Sentences, and include correct capitalization or punctuation for full credit. If Essay question requires a calculation you must complete the following:
* All set up and steps of work must be shown
* Answer must include correct sig figs (see notes on class website) and correct units
* All molar masses must be calculated with no rounding from the class periodic table.

Answer 1

Question 1: To find the number of grams of ammonium nitrate in 4.524 mols, we can use the molar mass of ammonium nitrate which is 80.043 g/mol. By multiplying the number of moles by the molar mass, we can determine that there are 362.23 grams of ammonium nitrate in 4.524 mols.

Question 2: In 2.15 mols of ammonium nitrate, there are 1.30 x 10^24 molecules of the compound. This is calculated by multiplying Avogadro's number (6.022 x 10^23) by the number of moles.

Question 3:
A. The full reaction for the decomposition of silver bromide to form solid silver and bromine gas is: 2AgBr(s) -> 2Ag(s) + Br2(g). This reaction is balanced with two moles of silver bromide producing two moles of silver and one mole of bromine gas.

B. To determine the number of grams of bromine gas produced when 25.54 grams of silver bromide decomposes fully, we first need to convert the given mass of silver bromide to moles using the molar mass of silver bromide (187.77 g/mol). Once we have the number of moles of silver bromide, we can use the stoichiometry of the balanced chemical equation to find the number of moles of bromine gas produced. Finally, we convert the moles of bromine gas to grams using the molar mass of bromine gas (79.904 g/mol). This calculation shows that 7.08 grams of bromine gas will be produced when 25.54 grams of silver bromide decomposes fully.