Find grams of HNO3: 6.5 moles * 63g/mole =409.5 g of HNO3 was used as the reactant!I hope it all makes sense!-1. 2NH3 + H2SO4 --> (NH4)2SO422.......
2. The moles are 4.54mol of C : 9.15mol of H and 2.27mol of O
2a: finding the ration by dividing with the lowest: C2H4O
3. Giving us C2H4O
4. MW of C2H4O is: 44g/mol so twice that is 88g/mol Thus, MW of 88amu is C4H8O2
Question 3
A is correct
B)
1. if 6.5 moles of NO2 is produced then 6.5 *4 moles of HNO3 was reacted because ration HNO3 to NO2 is 4:1 so you always have to look at the coefficient and relate the moles of the two compounds.
You can write it as follow:
6.5 moles NO2 * 4 moles of HNO3 / 1 mole of NO2
2. Now Find the MW of HNO3 : [1+ 14 + 3*16] g/moles = 63 g/mole
3. Find grams of HNO3 : 6.5 moles * 63g/mole = 409.5 g of HNO3 was used as the reactant!
I hope it all makes sense!
1. 2NH3 + H2SO4 --> (NH4)2SO4
22.7g NH3 / 17g/mole = 1.34moles NH3
54.8g H2SO4 / 98g/mole = 0.56moles H2SO4
1.34moles NH3 x (1H2SO4 / 2NH3) = 0.67moles H2SO4 required but we only have 0.56moles. H2SO4 = limiting reactant
0.56moles H2SO4 x (1(NH4)2SO4 / 2NH3) x 132g/mole = 36.96g produced
2. since the sum of the % must = 100%, we can say that we have 100g of the substance and the % of each element = mass of each element in the compound.
54.33g C / 12g/mole = 4.53moles C
9.15g H / 1g/mole = 9.15moles H
36.32g O / 16g/mole = 2.27moles O
molar ratios:
4.53C / 2.27 O = 2C
9.15H / 2.27 O = 4H
2.27 O / 2.27 O = 1 O
formula = C2H4O = 44g/mole. for 88g/mole, the molecular formula = C4H8O2
3. 4HNO3 --> 4NO2 + 2H2O + O2
6.5moles NO2 x (4HNO3 / 4NO2) x 63g/mole = 409.5g HNO3 reacted
