4NH3 + 7O2 = 4NO2 + 6H2O HHOW MANY MOLES OF AMMONIA WILL BE REQUIRED TO PRODUCE 10.0 MOLES OF WATER?
The equation tells you 4 mols NH3 will produce 6 mols H2O. Can you set up a proportion to make that work? For example, 2 mols NH3 would produce 3 mols H2O and 8 mols NH3 would produce 12 mols H2O.
twretrew
To set up a proportion, you can use the ratio from the balanced equation: 4 moles of NH3 will produce 6 moles of H2O.
Let's set up the proportion:
(4 moles NH3 / 6 moles H2O) = (x moles NH3 / 10 moles H2O)
To find the number of moles of NH3 required to produce 10.0 moles of water (H2O), we will solve for x (the unknown number of moles of NH3).
Cross-multiplying the equation gives us:
4 moles NH3 * 10 moles H2O = 6 moles H2O * x moles NH3
40 moles NH3 = 6x moles NH3
Dividing both sides by 6 gives us:
40 moles NH3 / 6 = x moles NH3
Simplifying further:
6.67 moles NH3 = x moles NH3
Therefore, approximately 6.67 moles of ammonia (NH3) would be required to produce 10.0 moles of water (H2O) according to the given equation.