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max2010maxim [7]
3 years ago
13

A student performed a chemical reaction in which 35 grams of hydrogen and 65 grams of oxygen reacted to form water. What is the

mass of the products? (answer with a number)
Chemistry
1 answer:
sergejj [24]3 years ago
8 0

Answer:

The mass of the product is 73,06 g

Explanation:

The reaction to form water is:

2H2 + O2 ---> 2H2O

35 g     65 g

Molar mass H2 = 2g/m ---> I have 17,5 moles (35g / 2 g/m)

Molar mass O2 = 32g/m ---> I have 2,03 moles (65g / 32g/m)

2 moles of H2 __reacts with __ 1 mol O2

17,5 moles of H2 __ reacts with ___ (17,5 m . 1 m) / 2 m = 8,75 m

I have just 2,03 m of O2 and I need 8,75 m so O2 is my limiting reagent

1 mol of O2 ___ reacts with 2 moles H2

2,03 moles of O2 __ reacts with  (2,03 m . 2 m) / 1 m = 4,06 m

I have 17,5 moles of H2 and I need 4,06 m. H2 is my excess reagent

<u><em>"All operations are done with the limiting reagent"</em></u>

1 mol O2 are required____ to form 2 H2O

2,03 mol O2 are required __ to form (2,03m . 2m) / 1m = 4,06 m

Molar mass of water: 18 g/m

Mass of water ----> Molar mass . Moles water = 18 g/m . 4,06 m = 73,08g

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2 years ago
2. (2 pts) How would you prepare 1.5 liters of 2 M KCI (MW=74.55 g/mol)
ra1l [238]

Answer:

Dissolve 226 g of KCl in enough water to make 1.5 L of solution

Explanation:

1. Calculate the moles of KCl needed

n = \text{1.5 L} \times \dfrac{\text{2 mol}}{\text{1 L}}= \text{3.0 mol}

2. Calculate the mass of KCl

m = \text{3.0 mol} \times \dfrac{\text{74.55 g}}{\text{1 mol}}= \text{224 g}

3. Prepare the solution

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Assuming that the reactants are:

(NH4)2SO4 (aq) + Ba(NO3)2 (aq)

and the products are:

BaSO4 (s) + 2NH4NO3 (aq),

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According to the solubility rules, the following elements are considered insoluble when paired with SO4:

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Therefore, the precipitate will be BaSO4 (s).

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