Lead (IV) oxide reacts with concentrated hydrochloric acid as follows: PbO2(s) + 4HCl(aq) -----> PbCl2(s) + Cl2(g) + 2H2O(l) What mass of lead chloride would be obtained from 37.2g of PbO2, and what mass of chlorine gas would be produced

In this question we have to use the equation Moles= Mass(g)/ Mr. We have the mass of PbO2 so to find the moles we need to divide the mass, which is 37.2g, by Its Mr, which is 207.2 + (162) = 239.2. the answer we get will be 37.2/239.2 = 0.1555 moles. Next we need to find the ratio between the moles of lead oxide and lead chloride, which we find to be a 1:1 therefore the number of moles of lead chloride is the same as lead oxide 0.1555 moles. To get the mass of lead chloride we have to times the moles by the Mr ( 207.2 + (35.52)) so 0.1555278.2= 43.3 g rounded to 3 sig fig.The second part of the question is to find the mass of chlorine. we have to find the ratio of moles between lead oxide and chlorine which we find to be 1:1 therefore the number of moles chlolrine is 0.1555 moles. to find the mass we have to multipy the moles by its Mr (35.52) so we get 0.1555*71=11.0g of chlorine gas.