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

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.