Which of the following would be considered a solution

Nitric acid, a major industrial and laboratory acid, is produced commercially by the multi-step Ostwald process, which begins wi

Marizza181 [45]

Answer: 3.178 Kg of ammonia must be used to produce 7.839 kg of nitric acid and The equations are not balanced. Explanation: Ostwald process is a multi-step for manufacturing Nitric Acid. First, We must check if our equations are balanced or not, by checking the amount of each element before and after the arrow. Step 1: NH3(g) + O2(g) ⟶ NO(g) + H2O(l) Step 2: NO(g) + O2(g) ⟶ NO2(g) Step 3: NO2(g) + H2O(l) ⟶ HNO3(l) + NO(g) For example in Step 1, the amount of Hydrogen atoms are different on both sides. On left side we can count 3 hydrogen atoms while on the right side we can count only 2 hydrogen atoms. In the Step 2, the amount of Oxygen atoms are different on both sides too. On the Left side we can count 3 oxygen atoms while on the right side we can count only 2 oxygen atoms and in step 3 the amount of Nitrogen atoms are different on both sides too. On the left side we can count 1 nitrogen atom while on the right side we can count 2 nitrogen atoms. So, Let´s balance equations by inspection, just adding coefficients to each compound, until the amount of atoms are equal on both sides. Step 1: 4NH3(g) + 5O2(g) ⟶4 NO(g) +6 H2O(l) Step 2: 2NO(g) + O2(g) ⟶ 2NO2(g) Step 3: 3NO2(g) + H2O(l) ⟶ 2HNO3(l) + NO(g) Second we gather the information what we are going to use in our calculations. Final Mass of HNO3 = 7,839Kg and Molecular Weigth = 63,01g/mol NH3 Molecular Weight= 17,031 g/mol Third we start discoverying the amount of NH3 that reacted completed to generate 7,839Kg of HNO3, using the giving equation and its respective molecular weights. 7,839Kg of HNO3 x 1000g / 1Kg x 1mol HNO3 / 63,01g HNO3 x 3mol NO2 / 2 mol HNO3 x 2 mol NO/ 2 mol NO2 x 4 mol NH3/4 mol NO x 17,031g NH3/1 mol NH3 x 1 Kg / 1000g= 3,178 Kg NH3 The amount of NH3 that is required to produce 7,839 Kg of HNO3 is 3,178 Kg NH3

7 0

3 years ago

Someone answer this please.

Nataly_w [17]

Answer: The percent composition of oxygen is 54.5%

Explanation:

Below I have attached my work through, but essentially the percent composition relies on finding the mass of the oxygens and dividing it by the mass of the whole compound. The masses of the elements can be found on the periodic table and added together to get a mass of the whole compound- ascorbic acid. Then, you turn this proportion into a percent by multiplying by 100%.

6 0

3 years ago

Read 2 more answers

A student performs an experiment to determine the density of a sugar solution. She obtains the following results: 1.79 g/mL, 1.8

ivann1987 [24]

Answer:

her results has good precision

Explanation:

accuracy and precision

4 0

3 years ago

Which factors in an ecosystem provide food (plants, fish, and livestock) or inhibit agriculture (insects)?

Pani-rosa [81]

Answer:

biotic

Explanation:

7 0

3 years ago

Aqueous hydrobromic acid (HBr) will react with solid sodium hydroxide (NaOH) to produce aqueous sodium bromide (NaBr) and liquid

bogdanovich [222]

Explanation:

Aqueous hydrobromic acid (HBr) will react with solid sodium hydroxide (NaOH) to produce aqueous sodium bromide (NaBr) and liquid water (H₂O). They will react according to the following equation.

HBr + NaOH ---> NaBr + H₂O

0.81 g of HBr are mixed with 0.568 g of NaOH. We have to find the mass of NaBr that can be produced. To do that we have to find which of the reactants is limiting the reaction. First, we will convert their grams into moles using their molar masses.

molar mass of HBr = 80.91 g/mol

molar mass of NaOH = 40.00 g/mol

mass of HBr = 0.81 g

mass of NaOH = 0.568 g

moles of HBr = 0.81 g * 1 mol/(80.91 g)

moles of HBr = 0.0100 moles

moles of NaOH = 0.568 g * 1 mol/(40.00 g)

moles of NaOH = 0.0142 moles

HBr + NaOH ---> NaBr + H₂O

Now if we take a quick look at the coefficients of the reaction we will see that 1 mol of HBr will react with 1 mol of NaOH since both coefficients are 1. Then their molar ratio is 1 : 1. That also means that 0.0100 moles of HBr will only react with 0.0100 moles of NaOH, and we have mixed 0.0142 moles of it. So, NaOH is in excess and HBr is the limiting reagent.

1 mol of HBr : 1 mol of NaOH molar ratio

moles of NaOH = 0.0100 moles of HBr * 1 mol of NaOH/(1 mol of HBr)

moles of NaOH = 0.0100 moles < 0.0142 moles ----> NaOH is in excess

And now that we know that HBr is the limiting reagent we can find the number of moles of NaBr that will be produced by 0.0100 moles of HBr. And finally convert those moles into grams using the molar mass.

1 mol of HBr : 1 mol of NaBr molar ratio

moles of NaBr = 0.0100 moles of HBr * 1 mol of NaBr/(1 mol of HBr)

moles of NaBr = 0.0100 moles

molar mass of NaBr = 102.89 g/mol

mass of NaBr = 0.0100 moles * 102.89 g/mol

mass of NaBr = 1.0289 g

mass of NaBr = 1.0 g

Answer: the maximum mass of sodium bromide that could be produced is 1.0 g.

7 0

1 year ago