Answer:
To find the mass of water produced at STP (Standard Temperature and Pressure) when 0.671 L of nitrogen monoxide is produced, we need to use the stoichiometry of the balanced chemical equation for the reaction.
The balanced equation for the reaction of nitrogen monoxide (NO) with water (H2O) is:
2NO + O2 -> 2NO2
From this equation, we can see that 2 moles of NO produce 2 moles of NO2. Since the volume of NO is given in liters, we need to convert it to moles using the ideal gas law.
1 mole of any gas occupies 22.4 liters at STP. Therefore, we can calculate the number of moles of NO as follows:
0.671 L NO * (1 mole NO / 22.4 L NO) = 0.030 moles NO
Now, we can use the stoichiometry of the balanced equation to determine the number of moles of water produced. From the equation, we can see that for every 2 moles of NO, 2 moles of NO2 are produced. Therefore, 0.030 moles of NO will produce 0.030 moles of NO2.
Since the reaction is balanced, we know that for every mole of NO2 produced, one mole of water is produced. Therefore, the number of moles of water produced is also 0.030 moles.
To find the mass of water produced, we need to multiply the number of moles of water by its molar mass. The molar mass of water (H2O) is approximately 18.015 g/mol.
Mass of water produced = 0.030 moles * 18.015 g/mol = 0.540 g
Therefore, the mass of water produced at STP when 0.671 L of nitrogen monoxide is produced is 0.540 grams.
Step-by-step explanation: