1340.40 grams of Iron is made to react with nitric acid, find mass of ferri nitrite produced. Use the given equation 2 Fe + 6HNO3 2Fe (NO3 + 3H₂ ​

The mass of ferric nitrate, Fe(NO₃)₃ produced from the reaction given the data is 5792.44 g

2Fe + 6HNO₃ –> 2Fe(NO₃)₃ + 3H₂

Molar mass of Fe = 56 g/mole

Mass of Fe from the balanced equation = 2 × 56 = 112 g

Molar mass of Fe(NO₃)₃ = 56 + 3[14 + (16×3)] = 242 g/mole

Mass of Fe(NO₃)₃ from the balanced equation = 2 × 242 = 484 g

SUMMARY

From the balanced equation above,

112 g of Fe reacted to produce 484 g of Fe(NO₃)₃

From the balanced equation above,

112 g of Fe reacted to produce 484 g of Fe(NO₃)₃

Therefore,

1340.4 g of Fe will react to produce = (1340.4 × 484) / 112 = 5792.44 g of Fe(NO₃)₃

Thus, 5792.44 g of Fe(NO₃)₃ were obtained from the reaction

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PLEASE HELP!!
In the formula
1N2 +3H2 → 2NH3 + 1H2
1 diatomic nitrogen (1N2) and 3 diatomic hydrogen (3H2) produce 2 ammonia (2NH3) with 1 diatomic hydrogen (1H2) left over. How can you change the formula so no diatomic hydrogen (H2) is left over?

(1 point)

Remove a H2 molecule from the left side of the equation.


Remove a NH3 molecule from the right side of the equation.


There is nothing that can be done to use the leftover molecules.


Add another N2 molecule to the left side of the equation.