If you have 1.4 mol of carbon before the reaction you have 1,4 mol of carbon after the reaction. If you make carbon monoxide, CO, then you need 1.4 mol of CO after the reactiion. If you make carbon dioxide, CO2, you need 1.4 mol of C2O, because you have one carbon in each molecule.
14
If the reaction were 2AlCl3 + 3Pb ---> 3PbCl2 + 2Al, then from this balanced equation, 14 moles of AlCl3 would produce 3/2 x 14 moles of PbCl2 = 21 moles of PbCl2.
None, unless there is metallic potassium in the reaction mixture. Assuming excess potassium metal is present then 14 moles of KBr can be produced. 7BaBr2 + excess potassium -----> 14KBr + 7 Ba
Since 14 (4+10) moles of P4O10 contains 4 moles of Phosphorus, 8 moles of P4O10 will contain :: (8 x 4)/14 = 2.286 moles of Phosphorus
The answer is 14 moles.
14
If the reaction were 2AlCl3 + 3Pb ---> 3PbCl2 + 2Al, then from this balanced equation, 14 moles of AlCl3 would produce 3/2 x 14 moles of PbCl2 = 21 moles of PbCl2.
2KClO3 + heat -> 2KCl + 3O2 14 moles KClO3 (3 mole O2/2 mole KClO3) = 21 moles oxygen made This is a common industrial method of producing oxygen.
None, unless there is metallic potassium in the reaction mixture. Assuming excess potassium metal is present then 14 moles of KBr can be produced. 7BaBr2 + excess potassium -----> 14KBr + 7 Ba
Since 14 (4+10) moles of P4O10 contains 4 moles of Phosphorus, 8 moles of P4O10 will contain :: (8 x 4)/14 = 2.286 moles of Phosphorus
1,0.10e9 atoms is equivalent to 0,166.10e-14 moles.
The answer is 14 moles.
When methane undergoes complete combustion, the equation for the reaction is CH4 + 2 O2 -> CO2 + 2 H2O. This shows that the number of moles of carbon dioxide formed are the same as the number of moles of methane reacted, so that 14 moles of carbon dioxide will be formed from 14 moles of methane.
The calculation is: Moles = Mass / Atomic Mass Moles = 0.085 / 17 Moles = 0.005 Atomic mass is 17 because ammonia is NH3, with N = 14, and H = 1. 14 + 1 + 1 + 1 = 17.
Step 1: balanced equation for the synthesis of ammonia:N2 + 3H2 --> 2NH3Step 2: convert 10g H2 to moles:H2=2 g/mol; 10g = 5 moles of hydrogen.Step 3: use molar ratios (coefficients) in balanced equation to determine moles of product:3/2=5/x; solve for x to get 3.3 moles of ammonia (product).Step 4: use relationship between moles and grams to get grams of product:NH3=14+1+1+1=17 g/mol. Given 3.3 moles, 17(3.3)=56.1 grams NH3.
There are 29/14, or just over 2 moles of nitrogen in 19 grams.
1 000 000 000 molecules of H2O2 is equivalent to 0,166.10e-14 moles.