2.95 mole H2O (2 moles H/1 mole H2O)
= 5.90 moles hydrogen
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To form water (H₂O), the balanced chemical reaction is 2 H₂ + O₂ → 2 H₂O. This indicates that 2 moles of hydrogen gas (H₂) are required to produce 2 moles of water (H₂O). Therefore, for 8.12 moles of H₂O, you would need 8.12 moles of H₂, as the ratio of H₂ to H₂O is 1:1.
The answer is 10 moles.
The combustion of butane (C₄H₁₀) can be represented by the balanced equation: 2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O. From this equation, we see that 2 moles of butane produce 8 moles of carbon dioxide. Therefore, if 5.31 moles of C₄H₁₀ are used, the moles of CO₂ produced can be calculated as follows: (5.31 moles C₄H₁₀) × (8 moles CO₂ / 2 moles C₄H₁₀) = 21.24 moles of CO₂.
The combustion reaction of ethane (C2H6) with oxygen produces carbon dioxide and water. The balanced equation for the combustion of ethane is: [ 2 \text{C}_2\text{H}_6 + 7 \text{O}_2 \rightarrow 4 \text{CO}_2 + 6 \text{H}_2\text{O} ] From the equation, 2 moles of ethane produce 6 moles of water. Therefore, 4 moles of ethane will produce ( \frac{6}{2} \times 4 = 12 ) moles of water.
To determine the number of moles of water needed to react with 1.7 moles of Li2O, we can use the balanced chemical equation for the reaction: [ \text{Li}_2\text{O} + 2\text{H}_2\text{O} \rightarrow 2\text{LiOH}. ] From the equation, 1 mole of Li2O reacts with 2 moles of water. Therefore, 1.7 moles of Li2O would require 1.7 x 2 = 3.4 moles of water.
To form water (H₂O), the balanced chemical reaction is 2 H₂ + O₂ → 2 H₂O. This indicates that 2 moles of hydrogen gas (H₂) are required to produce 2 moles of water (H₂O). Therefore, for 8.12 moles of H₂O, you would need 8.12 moles of H₂, as the ratio of H₂ to H₂O is 1:1.
In 1 mole of water (H2O), there are 2 moles of hydrogen (H). This means that in 2.08 moles of water, there are 2.08 x 2 = 4.16 moles of hydrogen. To convert moles to grams, we use the molar mass of hydrogen: 4.16 moles x 1.01 g/mol = 4.22 grams of hydrogen.
The answer is 10 moles.
2H2 + O2 --> 2H2OAs you can see by the balanced reaction, for every 1 mole of oxygen used, 2 moles of water are formed. Also notice that for every 1 mole of oxygen used, you need 2 moles of hydrogen to produce the 2 moles of water. So in your case 110 moles of oxygen would produce 220 moles of water & would also require 220 moles of hydrogen (which you have in excess since you have 230 moles of hydrogen). So 220 moles of water are the most that can be formed.
2.5 moles H2O (2 moles H/1 mole H2O) = 5 moles of hydrogen
The combustion of butane (C₄H₁₀) can be represented by the balanced equation: 2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O. From this equation, we see that 2 moles of butane produce 8 moles of carbon dioxide. Therefore, if 5.31 moles of C₄H₁₀ are used, the moles of CO₂ produced can be calculated as follows: (5.31 moles C₄H₁₀) × (8 moles CO₂ / 2 moles C₄H₁₀) = 21.24 moles of CO₂.
1.5 moles of Hydrogen. In every mole of H2SO4 (Sulfuric Acid) there are 2 moles of Hydrogen atoms. So, in .75 moles of Sulfuric Acid, there would be 1.5 (double the moles of sulfuric acid) moles of Hydrogen.
Well if one mole of water = 2 moles of hydrogen and 1 mole of oxygen, than 2moles of water = 4 moles of hydrogen and 2moles of oxygen.
5.00 moles H x 1 mole C2H4O2/4 moles H = 1.25 moles of C2H4O2 present.
for each mole of anything there is 6.022x10^23 molecules. Therefore for 5 moles of water there is 5 x 6.022x10^23 = 3.011x10^24 molecules of water
The ratio H/O is 2.
The combustion reaction of ethane (C2H6) with oxygen produces carbon dioxide and water. The balanced equation for the combustion of ethane is: [ 2 \text{C}_2\text{H}_6 + 7 \text{O}_2 \rightarrow 4 \text{CO}_2 + 6 \text{H}_2\text{O} ] From the equation, 2 moles of ethane produce 6 moles of water. Therefore, 4 moles of ethane will produce ( \frac{6}{2} \times 4 = 12 ) moles of water.