When a 1.25 g sample of MgCO3·xH2O is heated to remove all water, a mass of 0.759 g remains. Determine the value of x?
2016-05-11 4:14 am
When a 1.25 g sample of MgCO3·xH2O is heated to remove all water, a mass of 0.759 g remains. Determine the value of x and report it in the answer box below. Only answers that are integers are allowed.
回答 (2)
2016-05-11 6:05 am
Molar mass of MgCO₃ = 24.3 + 12.0 + 16.0×3 = 84.3 g/mol
Molar mass of H₂O = 1.0×2 + 16.0 = 18.0 g/mol
In 1.25 g of MgCO₃•xH₂O :
Mass of MgCO₃ = 0.759 g
Mass of H₂O = 1.25 - 0.759 = 0.491
Mole ratio MgCO₃ : H₂O
= 0.759/84.3 : 0.491/18
= 0.009 : 0.027
= 1 : 3
Hence, x = 3
Molar mass of H₂O = 1.0×2 + 16.0 = 18.0 g/mol
In 1.25 g of MgCO₃•xH₂O :
Mass of MgCO₃ = 0.759 g
Mass of H₂O = 1.25 - 0.759 = 0.491
Mole ratio MgCO₃ : H₂O
= 0.759/84.3 : 0.491/18
= 0.009 : 0.027
= 1 : 3
Hence, x = 3
2016-05-11 4:27 am
um. .4
收錄日期: 2021-04-18 14:51:58
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