2 moles C8H18 (18 moles H/1 mole C8H18)
= 36 moles of hydrogen
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∙ 11y ago2 moles C8H18 (18 moles H/1 mole C8H18)(6.022 X 10^23/1 mole H) = 2.2 X 10^25 atoms of hydrogen
Balance this combustion reaction first! 2C4H10 + 13O2 -> 8CO2 + 10H2O 0.86 moles C4H10 (13 moles O2/2 moles C4H10) = 5.6 moles of oxygen required ----------------------------------------
Na2CO32 * 2 = 4 moles sodium.===========================
You have :N2 + O2 -----> 2NOmoles NO = ( 2 mols N2 reacted ) ( 2 mol NO / mol N2 reacted )moles NO = 4 moles NOmass NO = ( 4 mol NO ) ( 30 g NO / mol NO ) = 120 g NO
14.17 mol BaBr2 has 2*14.17 mol Br in it, so 28.34 mol KBr can be produced (also 28.34 mol K is needed)
2 moles C8H18 (18 moles H/1 mole C8H18)(6.022 X 10^23/1 mole H) = 2.2 X 10^25 atoms of hydrogen
Balanced equation. 2C8H18 + 25O2 -> 16CO2 + 18H2O 5.5 moles C8H18 (25 moles O2/2 moles C8H18) = 68.75 moles O2 needed
Balanced equation: 2C8H18 + 25O2 ==> 16CO2 + 18H2Omoles of octane used: 325 g x 1 mole/114g = 2.85 moles octanemoles H2O produced: 18 moles H2O/2 moles C8H18 x 2.85 moles C8H18 = 25.65 moles H2O
Balance this combustion reaction first! 2C4H10 + 13O2 -> 8CO2 + 10H2O 0.86 moles C4H10 (13 moles O2/2 moles C4H10) = 5.6 moles of oxygen required ----------------------------------------
Na2CO32 * 2 = 4 moles sodium.===========================
You have :N2 + O2 -----> 2NOmoles NO = ( 2 mols N2 reacted ) ( 2 mol NO / mol N2 reacted )moles NO = 4 moles NOmass NO = ( 4 mol NO ) ( 30 g NO / mol NO ) = 120 g NO
First write a balanced chemical equation: 2K + Br2 ---> 2KBR Find the limiting reactant by using the moles of each element and determining which one gives you the smallest number of moles of potassium bromide. 2.92 mol K (2 mol KBr/2 mol K)= 2.92 mol KBr 1.78 mol Br2 (2 mol KBR/1 mol Br2)=3.56 mol KBr potassium is your limiting reactant so the max. number of moles of KBr that can be produced is 2.92 mol of KBr
14.17 mol BaBr2 has 2*14.17 mol Br in it, so 28.34 mol KBr can be produced (also 28.34 mol K is needed)
14.17 mol BaBr2 has 2*14.17 mol Br in it, so 28.34 mol KBr can be produced (also 28.34 mol K is needed)
2 of octane react with 25 of oxygen so 12.5 moles of O2 react with 1 mole of C8H18.
0.5 Moles If you have a 0.25 M solution, you have 0.25 mol/dm3, or 0.25 moles in 1 L (0.25 mol/L) If you have 2 L of solution, you have 2 L x 0.25 mol/L = 0.5 mol The L's cancel out, and you're left with moles.
2.8 moles Ca3N2 (3 mole Ca 2+/1 mole Ca3N2) = 8.4 moles Ca 2+ =============