Multiple Choice
How many moles of sodium carbonate (Na2CO3) contain 1.773 × 10^{17} carbon atoms?
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2. Atoms & Elements
Mole Concept
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How many moles of MgCO_3 are present in 252.939 grams of MgCO_3?
A
4.00 moles
B
0.50 moles
C
2.50 moles
D
1.50 moles
Verified step by step guidance1
Identify the given information: the mass of magnesium carbonate (MgCO_3) is 252.939 grams, and you need to find the number of moles.
Calculate the molar mass of MgCO_3 by adding the atomic masses of its elements: magnesium (Mg), carbon (C), and oxygen (O). Use the atomic masses: Mg = 24.31 g/mol, C = 12.01 g/mol, and O = 16.00 g/mol. The formula is: \(\text{Molar mass of MgCO}_3 = 24.31 + 12.01 + (3 \times 16.00)\) g/mol.
Use the formula relating mass, moles, and molar mass: \(n = \frac{m}{M}\), where \(n\) is the number of moles, \(m\) is the mass in grams, and \(M\) is the molar mass in g/mol.
Substitute the given mass (252.939 g) and the calculated molar mass into the formula: \(n = \frac{252.939}{\text{Molar mass of MgCO}_3}\).
Perform the division to find the number of moles of MgCO_3 present in the sample.
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