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Hello!
For this reaction: 2H₂ + O₂ → 2H₂O
If 19 g of oxygen reacts completely, you'll need 2,39 g of H₂
To calculate that, you need to use the following conversion factor, applying molar equivalences, reaction coefficients, and molar masses, to go from grams of Oxygen to grams of Hydrogen:
[tex] H_{2} Mass=19g O_{2}* \frac{1 mol O_{2}}{31,9988gO_{2}}* \frac{2 mol H_{2}}{1 mol O_{2}}* \frac{2,01588 gH_{2}}{1 mol H_{2}} =2,39gH_{2} [/tex]
For this reaction: 2H₂ + O₂ → 2H₂O
If 19 g of oxygen reacts completely, you'll need 2,39 g of H₂
To calculate that, you need to use the following conversion factor, applying molar equivalences, reaction coefficients, and molar masses, to go from grams of Oxygen to grams of Hydrogen:
[tex] H_{2} Mass=19g O_{2}* \frac{1 mol O_{2}}{31,9988gO_{2}}* \frac{2 mol H_{2}}{1 mol O_{2}}* \frac{2,01588 gH_{2}}{1 mol H_{2}} =2,39gH_{2} [/tex]
Answer: 2.4 grams
Explanation:
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
For [tex]O_2[/tex]
Given mass = 19 g
Molar mass of [tex]O_2[/tex] = 32 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of}O_2=\frac{19}{32}=0.60moles[/tex]
[tex]2H_2+O_2\rightarrow 2H_2O[/tex]
1 mole of [tex]O_2[/tex] reacts with 2 moles of [tex]H_2[/tex]
0.60 moles of [tex]O_2[/tex] wil react with =[tex]\frac{2}{1}\times 0.60=1.20[/tex] moles of [tex]H_2[/tex]
mass of [tex]H_2=moles\times {\text {molar mass}}=1.20moles\times 2g/mol=2.4g[/tex]
Thus 2.4 grams of hydrogen will react completely with 19 g of oxygen.