How many moles of O is present in 34 grams of Mg(NO_(2))_(2) a. 1.17 b. 1.61 c. 4.00 d. 0.29 e. 0.58

To find the number of moles of oxygen (O) in 34 grams of $Mg(NO_{2})_{2}$, we need to first determine the molar mass of $Mg(NO_{2})_{2}$ and then calculate the moles of oxygen based on the given mass.<br /><br />The molar mass of $Mg(NO_{2})_{2}$ is calculated as follows:<br />- Magnesium (Mg) has a molar mass of approximately 24.31 g/mol.<br />- Nitrogen (N) has a molar mass of approximately 14.01 g/mol.<br />- Oxygen (O) has a molar mass of approximately 16.00 g/mol.<br /><br />The formula $Mg(NO_{2})_{2}$ indicates that there is one magnesium atom, two nitrogen atoms, and four oxygen atoms in each molecule. Therefore, the molar mass of $Mg(NO_{2})_{2}$ is:<br />$24.31 + 2(14.01 + 2(16.00)) = 24.31 + 2(46.01) = 24.31 + 92.02 = 116.33 \text{ g/mol}$<br /><br />Now, we can calculate the moles of $Mg(NO_{2})_{2}$ in 34 grams:<br />$\text{moles of } Mg(NO_{2})_{2} = \frac{34 \text{ g}}{116.33 \text{ g/mol}} \approx 0.292 \text{ moles}$<br /><br />Since there are four oxygen atoms in each molecule of $Mg(NO_{2})_{2}$, the moles of oxygen in 34 grams of $Mg(NO_{2})_{2}$ is:<br />$\text{moles of O} = 4 \times 0.292 = 1.168 \approx 1.17$<br /><br />Therefore, the correct answer is:<br />a. 1.17