Answer:
[tex]m_{H_2O}=73.0gH_2O[/tex]
Explanation:
Hello there!
In this case, since the formation of water from hydrogen and oxygen is:
[tex]2H_2+O_2\rightarrow 2H_2O[/tex]
Whereas we find a 2:2 mole ratio of hydrogen to water. In such a way, by using the Avogadro's number, the aforementioned mole ratio and the molar mass of water (18.02 g/mol), we obtain the following grams of water product:
[tex]m_{H_2O}=2.44x10^{24}molec*\frac{1molH_2}{6.022x10^{23}molec}*\frac{2molH_2O}{2molH_2}*\frac{18.02gH_2O}{1molH_2O}\\\\ m_{H_2O}=73.0gH_2O[/tex]
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