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 and correct units
* All molar masses must be calculated with no rounding from the periodic table.

Answer 1

Question 1: To find the number of grams of ammonium nitrate in 4.524 mols of the compound, we first need to calculate the molar mass of NH4NO3, which is 80.043 grams/mol. Then, we can use the formula: grams = mols x molar mass. So, 4.524 mols x 80.043 g/mol = 361.886 grams of ammonium nitrate.

Question 2: To determine the number of molecules in 2.15 mols of ammonium nitrate, we can use Avogadro's number, which is 6.022 x 10^23 molecules/mol. Therefore, 2.15 mols x 6.022 x 10^23 molecules/mol = 1.2973 x 10^24 molecules of ammonium nitrate.

Question 3:
A. The decomposition of silver bromide can be represented by the balanced equation: AgBr(s) → Ag(s) + Br2(g)

B. To find the number of grams of bromine gas produced when 25.54 grams of silver bromide decomposes fully, we first need to find the molar mass of AgBr, which is 187.77 grams/mol. Next, we can use stoichiometry to find the number of moles of Br2, which will be equal to the moles of AgBr. Therefore, 25.54 grams / 187.77 g/mol = 0.1357 mols of Br2. Finally, we can convert this to grams by multiplying by the molar mass of Br2, which is 159.808 g/mol: 0.1357 mols x 159.808 g/mol = 21.69 grams of bromine gas produced.