120 cm\(^3\) of hydrogen were sparked with 60cm\(^3\) of oxygen at 110°C. What was the
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120 cm\(^3\) of hydrogen were sparked with 60cm\(^3\) of oxygen at 110°C. What was the volume of steam produced? The equation for the reaction is 2H\(_{2(g)}\) + O\(_{2(g)} \rightarrow\) 2H\(_2\)O\(_{(g)}\)
- A)30cm\(^3\)
- B)60cm\(^3\)
- C)90cm\(^3\)
- D)120cm\(^3\)
Correct Answer: D)120cm\(^3\)
Explanation
First, let's look at the balanced equation for the reaction:
\(2H_{2(g)} + O_{2(g)} \rightarrow 2H_2O_{(g)}\)
This equation tells us that 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of steam (water vapor).
Now, let's convert the given volumes of hydrogen and oxygen into moles using their molar volumes. The molar volume of an ideal gas at standard temperature (0
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