Consider the reaction:
3Ba(s)+N2(g)--->Ba3N2(s)
1)How many moles of barium metal react to produce 0.190 mol of barium nitride?
2)How many moles of nitrogen gas react?
Use the coefficients in the balanced equation to convert moles of one substance to moles of another..
0.190 mole Ba3N2 x (3 moles Ba/1 mole Ba3N2) = 0.190 x 3/1 = ?
ua;hdi
To answer the questions, we need to use the balanced chemical equation for the reaction:
3Ba(s) + N2(g) → Ba3N2(s)
1) To determine how many moles of barium metal react to produce 0.190 mol of barium nitride, we can use the mole ratio from the balanced equation. In this reaction, the coefficient of Ba is 3, meaning that 3 moles of Ba react to produce 1 mole of Ba3N2.
So, to find the number of moles of Ba that react, we can set up the following ratio:
3 moles Ba / 1 mole Ba3N2 = x moles Ba / 0.190 moles Ba3N2
Cross multiplying, we get:
3 moles Ba = 0.190 moles Ba3N2 * x moles Ba
x = (0.190 moles Ba3N2 * 3 moles Ba) / 1 mole Ba3N2
x = 0.570 moles Ba
Therefore, 0.570 moles of barium metal react to produce 0.190 mol of barium nitride.
2) To determine how many moles of nitrogen gas react, we can again use the mole ratio from the balanced equation. In this reaction, the coefficient of N2 is 1, meaning that 1 mole of N2 reacts to produce 1 mole of Ba3N2.
So, the number of moles of N2 that react is the same as the number of moles of Ba3N2 formed, which is 0.190 moles.
Therefore, 0.190 moles of nitrogen gas react.
To find the number of moles of barium metal that react, we need to use the stoichiometry of the balanced equation.
1) Stoichiometry is the molar relationship between different substances in a chemical reaction. In this case, the balanced equation shows that 3 moles of barium (Ba) react with 1 mole of nitrogen gas (N2) to produce 1 mole of barium nitride (Ba3N2).
Given that the reaction produced 0.190 mol of barium nitride (Ba3N2), we can use the stoichiometry to determine the amount of barium metal (Ba) that reacted.
Using the stoichiometric ratio from the balanced equation, we can set up a simple proportion:
3 mol Ba / 1 mol Ba3N2 = x mol Ba / 0.190 mol Ba3N2
Cross-multiplying and solving for x, the number of moles of barium metal, we get:
x = (3 mol Ba / 1 mol Ba3N2) * (0.190 mol Ba3N2)= 0.570 mol Ba
Therefore, 0.570 moles of barium metal react to produce 0.190 mol of barium nitride.
2) Similarly, to find the number of moles of nitrogen gas that react, we can use the stoichiometry of the balanced equation.
From the equation, we know that 3 moles of barium (Ba) react with 1 mole of nitrogen gas (N2) to produce 1 mole of barium nitride (Ba3N2).
Using this information, we can set up a proportion:
1 mol N2 / 3 mol Ba = x mol N2 / 0.570 mol Ba
Cross-multiplying and solving for x, the number of moles of nitrogen gas, we get:
x = (1 mol N2 / 3 mol Ba) * (0.570 mol Ba) = 0.190 mol N2
Therefore, 0.190 moles of nitrogen gas react in this reaction.