In the flowing equitation 2H2+O2→2H2O how many grams of oxygen are needed to make 9g of water?

2H₂+O₂ ->2H₂OThe best way to answer any mass balancing questions like this is to set the information in a table as follows and fill in the blanks: 2H₂ + O₂ -> 2H₂OMass ? 9gMr Moles 1. The first step is to calculate the M_{r (}relative molecular mass) of oxygen and water To do this we need to use the atomic mass Hydrogen and oxygen (H=1 and O=16) 2H₂ + O₂ -> 2H₂OMass ? 9gMr 16+16=32 1+1+16=18 Moles 2. The second step is calculate the number of moles of water in 9gWe need to use the equation moles=mass/M_r 2H₂ + O₂ -> 2H₂OMass ? 9gMr 32 18 Moles 9/18=0.53. We can now work out the number of moles of oxygen needed to make 0.5 moles of water. To do this we look at the equation and see the ratio of oxygen to water is 1:2 (One molecule of oxygen reacts with two molecules of hydrogen to form 1 molecule of water). So should have half as many moles of oxygen as water. 2H₂ + O₂ -> 2H₂OMass ? 9gMr 32 18 Moles 0.5/2=0.25 0.54. Finally we can calculate the mass of oxygen needed. We need to rearrange the equation moles=mass/M_r to give: mass=molesxM_r 2H₂ + O₂ -> 2H₂OMass 0.25x32=8g 9gMr 32 18 Moles 0.25 0.5The answer is 8g of oxygen!