30gr .of me burnt with 32gr of O2 to produce Mgo. Find limiting reagent and calculate how much other reactant present in excess amount
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Please find below the solution of your asked query:
2H2 + O2 → 2H2O
According to the balanced chemical equation;
(2 * 2) = 4g of Hydrogen reacts with 32g of Oxygen.
So, 3g of Hydrogen will react with (3 * 32) / 4 = 24g of Oxygen.
Since, the amount of Oxygen given (29g) is more than the required amount of Oxygen (24g). Thus, oxygen is the excess reagent and hydrogen is the limiting reagent. The amount of water formed will depend upon the amount of Hydrogen.
4g of Hydrogen produces (2 * 18) g of water.
So, 3g of Hydrogen produces (2 * 18 * 3) / 4 = 27g of water
The amount of Oxygen left un-reacted = 29g - 24g = 5g
Mark it as brainly answer
Please find below the solution of your asked query:
2H2 + O2 → 2H2O
According to the balanced chemical equation;
(2 * 2) = 4g of Hydrogen reacts with 32g of Oxygen.
So, 3g of Hydrogen will react with (3 * 32) / 4 = 24g of Oxygen.
Since, the amount of Oxygen given (29g) is more than the required amount of Oxygen (24g). Thus, oxygen is the excess reagent and hydrogen is the limiting reagent. The amount of water formed will depend upon the amount of Hydrogen.
4g of Hydrogen produces (2 * 18) g of water.
So, 3g of Hydrogen produces (2 * 18 * 3) / 4 = 27g of water
The amount of Oxygen left un-reacted = 29g - 24g = 5g
Mark it as brainly answer
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