Please help! Urgent!

Small amounts of chlorine gas can be generated in the laboratory from the reaction of Manganese(IV) oxide with hydrochloric acid: 4HCl(aq) + MnO2(s) = 2H2O(l) + MnCl2(s) + Cl2(g). (a) What mass of Cl2 can be produced from 42.7 g of MnO2 with an excess of HCl(aq)? (b) What volume of chlorine gas (of density 3.17 g/L) will be produced from the reaction of 300 mL of .100M HCl(aq) with an excess of MnO2? (c) Suppose only 150mL of chlorine was produced in the reaction in part (b). What is the percentage yield of the reaction?

Please please please show work! Thank you very much!

4HCl + MnO2 --> 2H2O + MnCl2 + Cl2

a) 42.7g MnO2 / 87g/mole = 0.49moles MnO2
0.49moles MnO2 x (1Cl2 / 1MnO2) x 70.9g/mole = 34.8g Cl2 produced

b) 0.3L x 0.1M HCl = 0.03moles HCl
0.03moles HCl x (1Cl2 / 4HCl) x 70.9g/mole = 0.53g Cl2
0.53g Cl2 / 3.17g/ml = 0.167L or 167ml

c)150ml / 167ml x 100% = 89.8% yield

In part (c), as long as the two volumes are expressed in the same units, it doesn't matter whether you use mL or L.