Question

7H2O2 + N2H4 → 2HNO3 + 8H2O ; (H=1; N=14; O=16) 1. Berapa mol HNO3 yang terbentuk dari 0,025 mol N2H4? 2. Berapa mol H2O2 yang dibutuhkan jika 1,35 mol H2O yang terbentuk? 3. Berapa gram H2O2 yang dibutuhkan untuk menghasilkan 45, 8 g HNO3?

Answer 1

`_7H_(2)O_(2)+N_(2)H_(4)=_2HNO_(3)+_8H_(2)O` (H = 1; N = 14; O = 16) 1. How many moles of
`HNO_(3)` are formed from 0.025 moles of
`N_(2)H_(4)`? 2. How many moles of
`H_(2)O_(2)` are needed if 1.35 moles of
`H_(2)O` are formed? 3. How many grams of
`H_(2)O_(2)` are needed to produce 45, 8 g of
`HNO_(3)`?

1. 0.05 moles of
`HNO_(3)`

2. 1.18 moles of
`H_(2)O_(2)`

3. 86.5g of
`H_(2)O_(2)`

Step-by-step explanation:

First write the balanced chemical equation given by the problem:

`_7H_(2)O_(2)+N_(2)H_(4)=_2HNO_(3)+_8H_(2)O`

1. Calculate how many moles of
`HNO_(3)` are formed from 0.025 moles of
`N_(2)H_(4)` using the reaction stoichiometry:

`0.025molesN_(2)H_(4)*(2molesHNO_(3))/(1molN_(2)H_(4))=0.05molesHNO_(3)`

2. Calculate how many moles of
`H_(2)O_(2)` are needed if 1.35 moles of
`H_(2)O` are formed using the reaction stoichiometry:

`1.35molesH_(2)O*(7molesH_(2)O_(2))/(8molesH_(2)O)=1.18molesH_(2)O_(2)`

3. Calculate how many grams of
`H_(2)O_(2)` are needed to produce 45,8 g of
`HNO_(3)`:

- Calculate the molar mass of the
`H_(2)O_(2)`:

Molar mass H: 1
`(g)/(mol)`

Molar mass O: 16
`(g)/(mol)`

Molar mass
`H_(2)O_(2)`=(2*1
`(g)/(mol)`)+(2* 16
`(g)/(mol)`)=34
`(g)/(mol)`

- Calculate the molar mass of the
`HNO_(3)`:

Molar mass H: 1
`(g)/(mol)`

Molar mass N: 14
`(g)/(mol)`

Molar mass O: 16
`(g)/(mol)`

Molar mass
`HNO_(3)`=(1*1
`(g)/(mol)`)+(1*14
`(g)/(mol)`)+(3*16
`(g)/(mol)`)=63
`(g)/(mol)`

`45.8gHNO_(3)*(1molHNO_(3))/(63gHNO_(3))*(7molesH_(2)O_(2))/(2molesHNO_(3))*(34gH_(2)O_(2))/(1molH_(2)O_(2))=86.5gH_(2)O_(2)`