Answer:
[tex]_7H_{2}O_{2}+N_{2}H_{4}=_2HNO_{3}+_8H_{2}O[/tex] (H = 1; N = 14; O = 16) 1. How many moles of [tex]HNO_{3}[/tex] are formed from 0.025 moles of [tex]N_{2}H_{4}[/tex]? 2. How many moles of [tex]H_{2}O_{2}[/tex] are needed if 1.35 moles of [tex]H_{2}O[/tex] are formed? 3. How many grams of [tex]H_{2}O_{2}[/tex] are needed to produce 45, 8 g of [tex]HNO_{3}[/tex]?
1. 0.05 moles of [tex]HNO_{3}[/tex]
2. 1.18 moles of [tex]H_{2}O_{2}[/tex]
3. 86.5g of [tex]H_{2}O_{2}[/tex]
Explanation:
First write the balanced chemical equation given by the problem:
[tex]_7H_{2}O_{2}+N_{2}H_{4}=_2HNO_{3}+_8H_{2}O[/tex]
1. Calculate how many moles of [tex]HNO_{3}[/tex] are formed from 0.025 moles of [tex]N_{2}H_{4}[/tex] using the reaction stoichiometry:
[tex]0.025molesN_{2}H_{4}*\frac{2molesHNO_{3}}{1molN_{2}H_{4}}=0.05molesHNO_{3}[/tex]
2. Calculate how many moles of [tex]H_{2}O_{2}[/tex] are needed if 1.35 moles of [tex]H_{2}O[/tex] are formed using the reaction stoichiometry:
[tex]1.35molesH_{2}O*\frac{7molesH_{2}O_{2}}{8molesH_{2}O}=1.18molesH_{2}O_{2}[/tex]
3. Calculate how many grams of [tex]H_{2}O_{2}[/tex] are needed to produce 45,8 g of [tex]HNO_{3}[/tex]:
- Calculate the molar mass of the [tex]H_{2}O_{2}[/tex]:
Molar mass H: 1[tex]\frac{g}{mol}[/tex]
Molar mass O: 16[tex]\frac{g}{mol}[/tex]
Molar mass [tex]H_{2}O_{2}[/tex]=(2*1[tex]\frac{g}{mol}[/tex])+(2* 16[tex]\frac{g}{mol}[/tex])=34[tex]\frac{g}{mol}[/tex]
- Calculate the molar mass of the [tex]HNO_{3}[/tex]:
Molar mass H: 1[tex]\frac{g}{mol}[/tex]
Molar mass N: 14[tex]\frac{g}{mol}[/tex]
Molar mass O: 16[tex]\frac{g}{mol}[/tex]
Molar mass [tex]HNO_{3}[/tex]=(1*1[tex]\frac{g}{mol}[/tex])+(1*14[tex]\frac{g}{mol}[/tex])+(3*16[tex]\frac{g}{mol}[/tex])=63[tex]\frac{g}{mol}[/tex]
[tex]45.8gHNO_{3}*\frac{1molHNO_{3}}{63gHNO_{3}}*\frac{7molesH_{2}O_{2}}{2molesHNO_{3}}*\frac{34gH_{2}O_{2}}{1molH_{2}O_{2}}=86.5gH_{2}O_{2}[/tex]
