2H_(2)O_(2)+1N_(2)H_(4)arrow 1N_(2)+4H_(2)O How many moles of N_(2)H_(4) will be needed to completely react with 75.3 grams of H_(2)O_(2) 1280 moles 1.11 moles 19.4 moles 37.7 moles

To solve this problem, we need to find the number of moles of $N_{2}H_{4}$ needed to completely react with 75.3 grams of $H_{2}O_{2}$.<br /><br />Given the balanced chemical equation:<br />$2H_{2}O_{2}+1N_{2}H_{4}\rightarrow 1N_{2}+4H_{2}O$<br /><br />First, we need to calculate the molar mass of $H_{2}O_{2}$:<br />Molar mass of $H_{2}O_{2}$ = (2 * 1.01) + 16.00 = 34.02 g/mol<br /><br />Next, we can calculate the number of moles of $H_{2}O_{2}$:<br />Number of moles of $H_{2}O_{2}$ = 75.3 g / 34.02 g/mol = 2.22 moles<br /><br />According to the balanced chemical equation, 2 moles of $H_{2}O_{2}$ react with 1 mole of $N_{2}H_{4}$. Therefore, we can calculate the number of moles of $N_{2}H_{4}$ needed:<br />Number of moles of $N_{2}H_{4}$ = 2.22 moles of $H_{2}O_{2}$ * (1 mole of $N_{2}H_{4}$ / 2 moles of $H_{2}O_{2}$) = 1.11 moles<br /><br />Therefore, the correct answer is 1.11 moles.