MnO2(S)+HCI(aq)-> CI2(g)+MnCI2(aq)+ H2O(I) How many moles of HCI are required to produce 1.50kg MnCI2?
To determine the number of moles of HCI required to produce 1.50kg of MnCI2, we need to use the molar mass of MnCI2 to convert from mass to moles.
The molar mass of MnCI2 is:
M(Mn) = 54.94 g/mol
M(Cl) = 35.45 g/mol
M(MnCI2) = M(Mn) + 2 * M(Cl) = 54.94 g/mol + 2 * 35.45 g/mol = 125.84 g/mol
Now, we can calculate the number of moles of MnCI2:
n(MnCI2) = mass/Molar mass = 1.50kg / (125.84 g/mol) = 11.92 mol
According to the balanced chemical equation, the mole ratio between HCI and MnCI2 is 1:1. Therefore, the number of moles of HCI required is also 11.92 mol.