4NH_(3)+5O_(2)-gt 4NO+6H_(2)O How many grams of NO are formed if 6.30g of ammonia react with 1.80g of oxygen? 0.37g 0.045 1.35g 11.19

To determine how many grams of NO are formed when 6.30g of ammonia react with 1.80g of oxygen, we need to follow these steps:<br /><br />1. Calculate the molar masses of the reactants and products.<br />2. Convert the masses of the reactants to moles.<br />3. Determine the limiting reactant.<br />4. Use stoichiometry to find the mass of NO produced.<br /><br />### Step 1: Calculate Molar Masses<br /><br />- \( NH_3 \): \( 14.01 + 3 \times 1.01 = 17.04 \) g/mol<br />- \( O_2 \): \( 2 \times 16.00 = 32.00 \) g/mol<br />- \( NO \): \( 14.01 + 16.00 = 30.01 \) g/mol<br /><br />### Step 2: Convert Masses to Moles<br /><br />- Moles of \( NH_3 \): \( \frac{6.30 \text{ g}}{17.04 \text{ g/mol}} = 0.370 \text{ mol} \)<br />- Moles of \( O_2 \): \( \frac{1.80 \text{ g}}{32.00 \text{ g/mol}} = 0.05625 \text{ mol} \)<br /><br />### Step 3: Determine the Limiting Reactant<br /><br />The balanced equation is:<br />\[ 4NH_3 + 5O_2 \rightarrow 4NO + 6H_2O \]<br /><br />From the stoichiometry of the reaction:<br />- 4 moles of \( NH_3 \) react with 5 moles of \( O_2 \).<br />- Therefore, 1 mole of \( NH_3 \) reacts with \( \frac{5}{4} \) moles of \( O_2 \).<br /><br />Calculate the required moles of \( O_2 \) for the given moles of \( NH_3 \):<br />\[ 0.370 \text{ mol } NH_3 \times \frac{5}{4} = 0.4625 \text{ mol } O_2 \]<br /><br />Since we only have 0.05625 moles of \( O_2 \), \( O_2 \) is the limiting reactant.<br /><br />### Step 4: Use Stoichiometry to Find the Mass of NO Produced<br /><br />Using the limiting reactant (\( O_2 \)):<br />- According to the balanced equation, 5 moles of \( O_2 \) produce 4 moles of \( NO \).<br />- Therefore, 0.05625 moles of \( O_2 \) will produce:<br />\[ 0.05625 \text{ mol } O_2 \times \frac{4 \text{ mol } NO}{5 \text{ mol } O_2} = 0.045 \text{ mol } NO \]<br /><br />Now, convert the moles of \( NO \) to grams:<br />\[ 0.045 \text{ mol } NO \times 30.01 \text{ g/mol} = 1.35 \text{ g } NO \]<br /><br />Therefore, the mass of NO formed is \( 1.35 \text{ g} \).<br /><br />The correct answer is:<br />\[ 1.35 \text{ g} \]