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.
Answer:
Equation of Reaction
2AgNO3 + BaCl2 === 2AgCl + Ba(NO3)2
Molar Mass of AgNO3 = 170g/mol
Moles of reacting AgNO3 = 100g/170gmol-¹
=0.588moles of AgNO3
From the equation of reaction...2moles of AgNO3 reacts to Produce 2Moles of Silver Chloride
So Their ratio is 2:2.
This means that 0.588Moles of AgCl Will be produced too.
ANSWER...0.588MOLES OF AgCl WILL BE PRODUCED.
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