Question #70613
1. Balence the fallowing equation: Fe(s)+O2(g)----> Fe2O3(s)
2. Balence the fallowing equation: NH3(g)+O2(g)----> N2(g)+H20(g)
3. Hydrogen and oxygen react with each other according to the equation:
2 H2(g)+O2(g)---> 2 H2O(g)
If 5.00 mol of O2 react with an unlimited amount of hydrogen, what amount (in moles) of H2O will be produced?
4. A sample of helium gas has a valume of 620. mL at a temperature of 500. K. If we decrease the temperature to 100. K while keeping the pressure consistent, what will the new volume be?
2. Balence the fallowing equation: NH3(g)+O2(g)----> N2(g)+H20(g)
3. Hydrogen and oxygen react with each other according to the equation:
2 H2(g)+O2(g)---> 2 H2O(g)
If 5.00 mol of O2 react with an unlimited amount of hydrogen, what amount (in moles) of H2O will be produced?
4. A sample of helium gas has a valume of 620. mL at a temperature of 500. K. If we decrease the temperature to 100. K while keeping the pressure consistent, what will the new volume be?
Expert's answer
4Fe(s)+3O2(g) ----> 2Fe2O3(s)
4NH3(g)+3O2(g) ----> 2N2(g)+6H20(g)
As you can see, the coefficients in the reaction equation are equal for hydrogen and water. Thus, the number of the moles of water produced and hydrogen reacted must be the same – 5 mol.
According to the Charles’ law:
V/T=const
Then, the new volume will be:
V_2=(V_1 T_2)/T_1 =(620 mL·100K)/(500 K)=124 mL.
4NH3(g)+3O2(g) ----> 2N2(g)+6H20(g)
As you can see, the coefficients in the reaction equation are equal for hydrogen and water. Thus, the number of the moles of water produced and hydrogen reacted must be the same – 5 mol.
According to the Charles’ law:
V/T=const
Then, the new volume will be:
V_2=(V_1 T_2)/T_1 =(620 mL·100K)/(500 K)=124 mL.
