Answer:
57.6g
Explanation:
So, if in one mole of water, 16 g of oxygen atom is present. Then, in 3.6 moles of water, the mass of oxygen present will be 3.6×16=57.6g. Therefore, the amount of oxygen present in 3.6 g water is option (B)- 57.6 g.
Use the molar mass of ammonia to change the grams to moles and then use mole-mole ratio
100. g NH3 (1 mol NH3/ 17.04 g) (3 mol H2/ 2 mol NH)= 8.80 moles H2
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