1) Write the balanced equation to state the molar ratios:
<span>3H2(g) + N2(g) → 2NH3(g)
=> molar ratios = 3 mol H2 : 1 mol N2 : 2 mol NH3
What volume of nitrogen is needed to produce 250.0 L of ammonia gas at STP?
First, convert the 250.0 L of NH3 to number of moles at STP .
Use the fact that 1 mole of gas at STP occupies 22.4 L
=> 250.0 L * 1mol/22.4 L = 11.16 L
Second, use the molar ratio to find the number of moles of N2 that produces 11.16 L of NH3
=> 11.16 L NH3 * [1 mol N2 / 2 mol NH3] = 5.58 mol N2
Third, convert 5.58 mol N2 into liters at STP
=> 5.58 mol N2 * [22.4 L/mol] = 124.99 liters
Answer: 124,99 liters
What volume of hydrogen is needed to produce 2.50 mol NH3 at STP?
First, find the number of moles of H2 that produce 2.50 mol by using the molar ratios:
2.50 mol NH3 * [3mol H2 / 2 mol NH3] = 3.75 mol H2
Second, convert the number of moles to liters of gas at STP:
3.75 mol * 22.4 L/mol = 84 liters of H2
Answer: 84 liters
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Answer:
255.51cm3
Explanation:
Data obtained from the question include:
V1 (initial volume) =?
T1 (initial temperature) = 50°C = 50 + 273 = 323K
T2 (final temperature) = - 5°C = - 5 + 237 = 268K
V2 (final volume) = 212cm3
Using the Charles' law equation V1/T1 = V2/T2, the initial volume of the gas can be obtained as follow:
V1/T1 = V2/T2
V1/323 = 212/268
Cross multiply to express in linear form
V1 x 268 = 323 x 212
Divide both side by 268
V1 = (323 x 212)/268
V1 = 255.51cm3
Therefore, the initial volume of the gas is 255.51cm3
Answer:
it's food engineering obviously
Heat= mass * change in temperature* specific heat
specific heat=409 J/kg K
