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Arisa [49]
3 years ago
13

Calcium hydride (CaH2) reacts with water to form hydrogen gas: CaH2(s) + 2H2O(l) → Ca(OH)2(aq) + 2H2(g) How many grams of CaH2 a

re needed to generate 48.0 L of H2 gas at a pressure of 0.995 atm and a temperature of 32 °C?
Chemistry
1 answer:
pickupchik [31]3 years ago
7 0

Answer:

40.g CaH2

Explanation:

1. ideal gas law(PV = nRT) → use ideal gas law first when volume is given

P = 0.995atm

V = 48.0L H2

n = ?

R = 0.0821L atm/molK

T = 32 + 273 = 305K

n = (0.995atm x 48.0L H2)/(0.0821L atm/molK x 305K) → do not simplify as small decimals might change the answer

2. Conversions

2 H2 and 1 CaH2 → 1/2

(mole of H2) x 1/2 x (molar mass of CaH2)

(0.995atm x 48.0L H2)/(0.0821L atm/molK x 305K) x 1/2 x (40.08 + 2.02) = 40.g CaH2

the longer answer will be 40.14887882 but as the minimum sigfig given in the question is 2, it is 40.g CaH2.

Hope it helped!

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If the container is closed and the ethanol is allowed to reach equilibrium with its vapor, how many grams of liquid ethanol rema
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Explanation:

Let us assume that the given data is as follows.

        V = 3.10 L,        T = 19^{o}C = (19 + 273)K = 292 K

       P = 40 torr    (1 atm = 760 torr)

So,     P = \frac{40 torr}{760 torr} \times 1 atm

             = 0.053 atm

          n = ?

According to the ideal gas equation, PV = nRT.

Putting the given values into the above equation to calculate the value of n as follows.

                 PV = nRT

   0.053 atm \times 3.10 L = n \times 0.0821 L atm/mol K \times 292 K

                 0.1643 = n \times 23.97

                    n = 6.85 \times 10^{-3}

It is known that molar mass of ethanol is 46 g/mol. Hence, calculate its mass as follows.

               No. of moles = \frac{mass}{\text{molar mass}}

                 6.85 \times 10^{-3} = \frac{mass}{46 g/mol}  

                    mass = 315.1 \times 10^{-3} g

                              = 0.315 g

Thus, we can conclude that the mass of liquid ethanol is 0.315 g.

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3 years ago
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