3 years ago

If 852.04 g Al(NO3)3 reacted with the 741.93 g Na2CO3, which would be the limiting reagent?

Chemistry

1 answer:

Sedaia [141]3 years ago

6 0

Answer:

Al(NO3)3

Explanation:

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Given: $V_{1}$ = 2.0 mL, $P_{1}$ = 1.02 atm, $T_{1} = 22^{o}C$

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$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$

Substitute the values into above formula as follows.

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{1.02 atm \times 2.0 mL}{22^{o}C} = \frac{1.27 atm \times V_{1}}{100^{o}C}\\V_{1} = 7.3 mL$

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