Question 1(1 point) 2H_(2)(g)+O_(2)(g)arrow 2H_(2)O(g) A student wants to calculate the amount of water (H_(2)O) in grams produced from 3.5 moles of oxygen gas (O_(2)) 126 g b 0.229 63g d 3.949

## Step 1<br />The given chemical equation is \(2H_{2}(g) + O_{2}(g) \rightarrow 2H_{2}O(g)\). This equation tells us that 1 mole of oxygen gas (\(O_{2}\)) reacts with 2 moles of hydrogen gas (\(H_{2}\)) to produce 2 moles of water (\(H_{2}O\)).<br /><br />## Step 2<br />The molar mass of water (\(H_{2}O\)) is 18.01528 g/mol. This is the mass of one mole of water.<br /><br />## Step 3<br />The student has 3.5 moles of oxygen gas (\(O_{2}\)). According to the stoichiometry of the reaction, 1 mole of oxygen gas produces 2 moles of water. Therefore, 3.5 moles of oxygen gas will produce \(3.5 \times 2 = 7\) moles of water.<br /><br />## Step 4<br />To find the mass of water produced, we multiply the number of moles of water by the molar mass of water. So, the mass of water produced is \(7 \times 18.01528 = 126.10596\) g.