points (}_{2)O_(2)+N_(2)H_(4)arrow N_(2)+4H_(2)O .750 mol of N_(2)H_(4) reacts with excess H_(2)O_(2) how many moles of water will be produced? 1 mole 1.5 moles 3 moles 4 moles

To solve this problem, we need to use the stoichiometry of the balanced chemical equation.<br /><br />The balanced chemical equation is:<br />${}_{2}O_{2}+N_{2}H_{4}\rightarrow N_{2}+4H_{2}O$<br /><br />From the equation, we can see that 1 mole of $N_{2}H_{4}$ reacts with $H_{2}O_{2}$ to produce 4 moles of $H_{2}O$.<br /><br />Given that 0.750 mol of $N_{2}H_{4}$ reacts with excess $H_{2}O_{2}$, we can calculate the number of moles of water produced as follows:<br /><br />Number of moles of water produced = (4 moles of $H_{2}O$ / 1 mole of $N_{2}H_{4}$) × 0.750 mol of $N_{2}H_{4}$<br />Number of moles of water produced = 3 moles<br /><br />Therefore, the correct answer is 3 moles.