Question

A mixture was found to contain 1.05g of SiO2, 0.69g of cellulose and 1.82g of calcium carbonate. What percentage of calcium carbonate is in the mixture?

Answer 1

$1,05g+0,69g+1,82g=3,56g\\\\ 3,56g \ \ \ \Rightarrow \ \ \ 100\%\\ 1,82g \ \ \ \Rightarrow \ \ \ x\\\\ x=\frac{1,82g*100\%}{3,56g}\approx 51,12\%$

Answer 2

Answer:

51.1 % of the mixture is calcium carbonate

Explanation:

Step 1: Data given

A mixture contains 1.05 grams of SiO2, 0.69 grams of cellulose and 1.82 grams of CaCO3

Step 2: Calculate the total mass of the mixture

Total mass = mass of SiO2 + mass of cellulose + mass of CaCO3

Total mass = 1.05 grams + 0.69 grams + 1.82 grams

Total mass = 3.56 grams

Step 3: Calculate percentage of calcium carbonate is in the mixture

% caCO3 = (mass CaCO3 / mass of mixture) *100%

% CaCO3 = (1.82 grams / 3.56 grams ) * 100%

% CaCO3= 51.1 %

51.1 % of the mixture is calcium carbonate