The product is magnesium sulfate plus hydrogen gas,as per this equation:
<span>Mg + H2SO4 = MgSO4 + H<span>2(g)</span></span>
Assuming that the gas acts like an ideal gas, we can
calculate for the final volume using the ideal gas law:
PV = nRT
Where P = pressure, V = volume, n = number of moles, R = gas
constant, and T = temperature
Assuming that P, n, and R are constant throughout the
process, we can define another constant K:
V / T = K where
K = nR / P
Equating the initial and final states:
Vi / Ti = Vf / Tf
Substituting the given values:
11.5 cm^3 / 415 K = Vf / 200 K
Vf = 5.54 cm^3
Solution,
Mass of Mg - 21gm
molar mass of Mg - 24gm
moles = Given mass/ Molar mass
moles = 21/24 = 0.875
1 mole of Mg produce 1 mole of H2 gas
so 0.875 mole of Mg will produce 0.875 moles of H2 gas
One mole of H2 gas = 22.4 litre
0.875 mole of H2 gas = 0.875×22.4
0.875 mole of H2 gas = 19.60 litre
So the nearest ANSWER is Option four 19.37 litre.
N(C): N(H)=n(C): n(H)=6: 10
3×10²¹: x=6: 10
x=5×10²¹