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

How many moles of chromium(iii) nitrate are produced when chromium reacts with 0.85 moles of lead(iv) nitrate to produce chromiu

Alexxandr [17]

The moles of chromium (iii) nitrate produced is calculated as follows

write the equation for reaction

3 Pb(NO3)2 + 2 Cr = 2 Cr(NO3)3 + 3 Pb

by use of mole ratio between Pb(NO3)2 to Cr(NO3)3 which is 3 : 2 the moles of Cr(NO3)3 is therefore
= 0.85 x2 /3 = 0.57 moles

5 0

3 years ago

What is the mass of 1.56 X 10^24 formula units of Na2SO4?

Paladinen [302]

Answer:

$m=368 g$

Explanation:

Hello,

In this case, in order to compute the required mass, we first must notice that 6.022x10²³ formula units of sodium sulfate contain 1 mole of such compound (Avogadro's relationship). Moreover, one mole of sodium sulfate contains 142.04 g, which is in fact, the molar mass. Thereby, the required mass is computed via the following mole-mass-particles relationship:

$m=1.56x10^{24}f.u*\frac{1mol}{6.022x10^{23}f.u}*\frac{142.04g}{1mol} \\\\m=368 g$

Regards.

7 0

3 years ago

This begins as a nebula

PSYCHO15rus [73]

The answer is star...

5 0

3 years ago