2NaN3 --> 2Na + 3N2
2.88 mol of NaN3 reacted * (3 mol of N2 produced / 2 mol of NaN3 reacted) =
4.32 mol of N2
Balanced equation. 4Na + O2 -> 2Na2O 10 moles Na (2 moles Na2O/4 moles Na) = 5.0 moles Na2O produced
4Na+O2 2Na2O
two moles
Na +H2O -> NaOH +(1/2)H2 Every mole of Sodium requires one mole of water to make one mole of Sodium Hydroxide. So two moles of Sodium will produce two moles of Sodium Hydroxide. If there are three moles of water in the initial reaction then there will be one mole of water left over after reacting with two moles of Sodium. This reaction will produce half a mole of hydrogen gas.
.033 mol H2
0,75 moles of nitrogen
To calculate the total number of moles of sodium azide in a 52-g sample, you need to divide the mass of the sample by the molar mass of sodium azide (65.01 g/mol). So, 52 g / 65.01 g/mol = 0.799 moles of sodium azide.
Balanced equation. 4Na + O2 -> 2Na2O 10 moles Na (2 moles Na2O/4 moles Na) = 5.0 moles Na2O produced
The thermal decomposition reaction is:2 KO2------------K2O2 + O20,2 moles of O2 are produced from o,4 moles KO2.
Sodium reacts with water. 0.652 NaOH moles will form.
4Na+O2 2Na2O
Using the balanced chemical equation for the decomposition of ammonium nitrate, 2NH4NO3 --> 2N2O + O2 + 4H2O, we can see that 2 moles of nitrous oxide (N2O) is produced for every 2 moles of ammonium nitrate (NH4NO3). Therefore, 0.55 moles of NH4NO3 will produce 0.55 moles of N2O. The volume of N2O can then be calculated using the ideal gas law equation (V = nRT/P) and given conditions.
two moles
0,028 moles carbonic are obtained.
When 2 mol of Na2CO3 dissociate, it will produce 4 mol of Na+ ions and 2 mol of CO3^2- ions. Therefore, a total of 6 moles of ions are produced.
The number of atoms is 45,166.10e23.
Each molecule of C6H6 contains 6 carbon atoms, so when 1 mole of C6H6 decomposes, 6 moles of carbon atoms are obtained. Therefore, in a 1.68 mole sample of C6H6, 6 × 1.68 = 10.08 moles of carbon atoms can be obtained from the decomposition.