Question

A gold coin called a Krugerrand has approximately 31.1035 grams of gold in it. How many metric tons of 15 ppm grade gold ore need to be mined in order to get enough gold for a single Krugerrand? Show your work here. Remember that 1 metric ton = 1x10 Essay answers are limited to about 500 words (3800 characters maximum, including spaces) 3800 Characters) remaining

Answer 1

Given Data => 6 15 ppm grade ..(is parts per million) . That means 15 gm gold is found in 106C 1 million gram of ore) To find = a) 1 gold coin kougerpand - which weighs 31.1035 grams 6] Waste product Parta In question solution » is gm of gold 1 gmot gold tgm of gold - – → 1000000 grams 1000000 swams 66666.66 groms 1000 ou is grams We need 311 31.1035 grams 31.1035 X 1 gm -> 31.1035 X 66666.66 arom ( 1 gold coin) 31.1035gm -> 2073566.66 grams op 2.07 106 | 1 metric ton = 1x106 In so we need to mine 2.07 metric ton order to get 31.1035 grams of gold

Parte In the question, We get 2070 2073566.66 gang geams In which we will found 31.1035 gram of gold. So the waste product will be = = Total mined one - Gold found 2073 566.6666 - 31,1035 Total waste = 2073535.5631 gram product . As 1x106 we know I metric ton = 2.073 x 100 gram. which is 2007 metric tons 2.0735355631 to be approx