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Maurinko [17]
3 years ago
8

If 13.7 grams of manganese oxide reacts with excess hydrogen chloride gas, how many grams of water are formed?

Chemistry
1 answer:
Lerok [7]3 years ago
3 0

Answer:

5.67 g OF WATER WILL BE FORMED WHEN 13.7 g OF MnO2 REACTS WITH HCl GAS.

Explanation:

EQUATION FOR THE REACTION

Mn02 + 4HCl --------> MnCl2 + Cl2 + 2H2O

From the balanced reaction between manganese oxide and hydrogen chloride gas;

1 mole of MnO2 reacts to form 2 mole of water

At STP, the molecular mass of the sample is equal to the mole of the substance. So therefore:

(55 + 16 * 2) g of MnO2 reacts to form 2 * ( 1 *2 + 16) g of water

(55 + 32) g of MnO2 reacts to form 2 * 18 g of water

87 g of MnO2 reacts to form 36 g of water

If 13.7 g of MnO2 were to be used?

87 g of MnO2 = 36 g of H2O

13.7 g of MnO2 = ( 13.7 * 36 / 87) g of water

= 493.2 / 87 g of water

Mass of water = 5.669 g of water

Approximately 5.67 g of water will be formed when 13.7 g of manganese oxide reacts with excess hydrogen chloride gas.

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F. <em>None of the above </em>

<em>No O atoms are present</em> as reacting substances, only O_2 and H_2O molecules.

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We must use <em>oxidation numbers</em> to decide whether oxygen or water is the substance reduced.

The oxidation number of O changes from 0 in O_2 to -2 in OH^(-).

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